Tag: Trigonometric Equations

Given:
$\sin x = \dfrac{1}{2}\sqrt{3} = \sin 60^{\circ}.$
Possibility 1:
$x = 60^{\circ} + k \cdot 360^{\circ}$
For $k = 0$, we obtain $x = 60^{\circ}~~(\checkmark)$
For $k=1$, we obtain $x = 420^{\circ}~~(\text{X})$
Possibility 2:
$\begin{aligned} x & = (180-60)^{\circ} + k \cdot 360^{\circ} \\ x & = 120^{\circ} + k \cdot 360^{\circ} \end{aligned}$
For $k = 0$, we obtain $x = 120^{\circ}~~(\checkmark)$
For $k=1$, we obtain $x = 480^{\circ}~~(\text{X})$
Therefore, the values of $x$ satisfying the equation, expressed in set notation, are $\boxed{\{60^{\circ}, 120^{\circ}\}}.$
(Answer A)

Given:
$\cos x = \dfrac{1}{2}= \cos 60^{\circ}.$
Possibility 1:
$x = 60^{\circ} + k \cdot 360^{\circ}$
For $k = 0$, we obtain $x = 60^{\circ}~~(\checkmark)$
For $k=1$, we obtain $x = 420^{\circ}~~(\text{X})$
Possibility 2:
$x = -60^{\circ} + k \cdot 360^{\circ}$
For $k = 0$, we obtain $x = -60^{\circ}~~(\text{X})$
For $k=1$, we obtain $x = 300^{\circ}~~(\checkmark)$
For $k=2$, we obtain $x = 660^{\circ}~~(\text{X})$
Therefore, the values of $x$ satisfying the equation, expressed in set notation, are $\boxed{\{60^{\circ}, 300^{\circ}\}}.$
(Answer E)

Notice that the equation $\sqrt{2} \sin 3x = 1$ is equivalent to the equation
$\sin 3x = \dfrac{1}{\sqrt{2}} = \dfrac{1}{2}\sqrt{2} = \sin 45^{\circ}.$
Thus, we obtain the following 2 possibilities.
Possibility 1:
$\begin{aligned} 3x & = 45^{\circ} + k \cdot 360^{\circ} \\ & \text{Divide both sides by}~3 \\ x& = 15^{\circ} + k \cdot 120^{\circ} \end{aligned}$
If $k = 0$, we obtain $x = 15^{\circ}~~(\checkmark)$
If $k = 1$, we obtain $x = 135^{\circ}~~(\checkmark)$
If $k = 2$, we obtain $x = 255^{\circ}~~(\text{X})$
Possibility 2:
$\begin{aligned} 3x & = (180-45)^{\circ} + k \cdot 360^{\circ} \\ 3x & = 135^{\circ} + k \cdot 360^{\circ} \\ & \text{Divide both sides by}~3 \\ x & = 45 ^{\circ} + k \cdot 120^{\circ} \end{aligned}$
If $k = 0$, we obtain $x = 45 ^{\circ}~~(\checkmark)$
If $k = 1$, we obtain $x = 165^{\circ}~~(\checkmark)$
If $k = 2$, we obtain $x = 285^{\circ}~~(\text{X})$
Therefore, the solution set of the equation is
$\boxed{\{15^{\circ}, 45^{\circ}, 135^{\circ}, 165^{\circ}\}}.$
(Answer D)

Given:
$\begin{aligned} 2 \cos \left(x -\dfrac{\pi}{3}\right) & = \sqrt{3} \\ \cos \left(x-\dfrac{\pi}{3}\right) & = \dfrac{1}{2}\sqrt{3} \\ \cos \left(x -\dfrac{\pi}{3}\right) & = \cos \dfrac{\pi} {6} \end{aligned}$
Possibility 1:
$\begin{aligned} x -\dfrac{\pi} {3} & = \dfrac{\pi} {6} + k \cdot 2\pi \\ x & = \dfrac{\pi} {6} + \dfrac{\pi} {3} + k \cdot 2\pi \\ x & = \dfrac{\pi} {2} + k \cdot 2\pi \end{aligned}$
For $k = 0$, we obtain $x = \dfrac{\pi} {2}~~(\checkmark)$
For $k=1$, we obtain $x = 2\dfrac{1}{2}\pi~~(\text{X})$
Possibility 2:
$\begin{aligned} x -\dfrac{\pi} {3} & = -\dfrac{\pi} {6} + k \cdot 2\pi \\ x & = – \dfrac{\pi} {6} + \dfrac{\pi} {3} + k \cdot 2\pi \\ x & = \dfrac{\pi} {6} + k \cdot 2\pi \end{aligned}$
For $k = 0$, we obtain $x = \dfrac{\pi} {6}~~(\checkmark)$
For $k=1$, we obtain $x = 2\dfrac{1}{6}\pi~~(\text{X})$
Therefore, the solution set of the trigonometric equation is $\boxed{\left\{\dfrac{\pi} {6}, \dfrac{\pi} {2}\right\}}.$
(Answer A)

Given:
$\tan 2x = \dfrac{1}{3}\sqrt{3} = \tan 30^{\circ}.$
Thus, we write
$\begin{aligned} 2x & = 30^{\circ} + k \cdot 180^{\circ} \\ x & = 15^{\circ} + k \cdot 90^{\circ} \end{aligned}$
For $k = 0$, we obtain $x = 15^{\circ}~~(\checkmark)$
For $k=1$, we obtain $x = 105^{\circ}~~(\checkmark)$
For $k=2$, we obtain $x = 195^{\circ}~~(\checkmark)$
For $k=3$, we obtain $x = 285^{\circ}~~(\text{X})$
Therefore, the values of $x$ satisfying the equation, expressed in set notation, are $\boxed{\{15^{\circ}, 105^{\circ}, 195^{\circ}\}}.$
(Answer A)

Use the following identity.
$\boxed{\cos^2 x = 1 -\sin^2 x}$
Therefore, we can write
$\begin{aligned} 2 \cos^2 x + 5 \sin x -4 & = 0 \\ 2(1 -\sin^2 x) + 5 \sin x -4 & = 0 \\ -2 \sin^2 x + 5 \sin x -2 & = 0 \\ \text{Multiply both sides by}~&-1 \\ 2 \sin^2 x -5 \sin x + 2 & = 0 \\ \text{Let}~\sin x & = y \\ 2y^2 -5y + 2 & = 0 \\ (2y-1)(y-2) & = 0 \end{aligned}$
From here, we obtain
$2y-1=0 \Leftrightarrow y = \dfrac{1}{2}$
or
$y-2=0 \Leftrightarrow y = 2.$
Substitute back $y = \sin x$ so that we obtain the following possibilities:
Possibility 1:
$\begin{aligned} \sin x & = \dfrac{1}{2} = \sin 30^{\circ} \\ x & = 30^{\circ} + k(360^{\circ}) \\ x & = (180-30)^{\circ} + k(360^{\circ}) \end{aligned}$
For $k = 0$, we then obtain
$x = 30^{\circ}$ or $x = 150^{\circ}$
Possibility 2: $\sin x = 2$
Since the maximum value of sine is $1$, then $\sin x = 2$ has no solution.
Therefore, the solution set of the trigonometric equation is $\{30^{\circ}, 150^{\circ}\}.$
(Answer A)

Use the following identity.
$\boxed{\cos 2x = 1 -2 \sin^2 x}$
Therefore, we can write
$\begin{aligned} \cos 2x -\sin x & = 0 \\ (1-2 \sin^2 x) – \sin x & = 0 \\ -2 \sin^2 x -\sin x + 1 & = 0 \\ \text{Multiply both sides by}~&-1 \\ 2 \sin^2 x + \sin x- 1 & = 0 \\ (2 \sin x -1)(\sin x + 1) & = 0 \end{aligned}$
Zero factors:
$2 \sin x -1 = 0 \Leftrightarrow \sin x = \dfrac{1}{2}$
or
$\sin x + 1 = 0 \Leftrightarrow \sin x = -1.$
Possibility 1:
$$\begin{aligned} \sin x & = \dfrac{1}{2} = \sin 30^{\circ} \\ x & = 30^{\circ} + k(360^{\circ}) \lor x = (180-30)^{\circ} + k(360^{\circ}) \end{aligned}$$For $k = 0$, we then obtain $x = 30^{\circ}$ or $x = 150^{\circ}$
Possibility 2: $\sin x = -1$
The only value of $x$ satisfying $\sin x = -1$ is $x = 270^{\circ}$, but since it lies outside the interval for $x$, it is not included in the solution set.
Therefore, the solution set of the trigonometric equation is $\{30^{\circ}, 150^{\circ}\}.$
(Answer A)

Use the sum formula for sine functions.
$$\boxed{\sin A + \sin B = 2 \sin \dfrac{1}{2}(A+B) \cos \dfrac{1}{2}(A-B)}$$Therefore, we can write
$$\begin{aligned} \sin (2x + 110)^{\circ} + \sin (2x-10)^{\circ} & = \dfrac{1}{2} \\ 2 \sin \dfrac{1}{2}(2x+110+2x-10)^{\circ} \\ \cos \dfrac{1}{2}(2x+110-2x+10)^{\circ} & = \dfrac{1}{2} \\ 2 \sin (2x + 50)^{\circ} \color{blue}{\cos 60^{\circ}} & = \dfrac{1}{2} \\ \cancel{2} \sin(2x+50)^{\circ} \cdot \color{blue}{\dfrac{1}{\cancel{2}}} & = \dfrac{1}{2} \\ \sin(2x+50)^{\circ} & = \dfrac{1}{2} \\ \sin(2x+50)^{\circ} & = \sin 30^{\circ} \end{aligned}$$Use the sine-form equation: $\sin px = \sin \theta$, with the solutions given by
$\boxed{\begin{aligned} px & = \theta + k \cdot 360^{\circ} \\ px & = (180^{\circ} -\theta) + k \cdot 360^{\circ}. \end{aligned}}$
Possibility 1:
$\begin{aligned} (2x + 50)^{\circ}& = 30^{\circ} + k \cdot 360^{\circ} \\ 2x^{\circ} & = -20^{\circ} + k \cdot 360^{\circ} \\ x^{\circ} & = -10^{\circ} + k \cdot 180^{\circ} \end{aligned}$
Choose $k = 1$ or $k =2$ so that we obtain $x = 170^{\circ}$ or $x = 350^{\circ}.$
Possibility 2:
$\begin{aligned} (2x + 50)^{\circ}& = (180-30)^{\circ} + k \cdot 360^{\circ} \\ 2x^{\circ} & = 100^{\circ} + k \cdot 360^{\circ} \\ x^{\circ} & = 50^{\circ} + k \cdot 180^{\circ} \end{aligned}$
Choose $k = 0$ or $k =1$ so that we obtain $x = 50^{\circ}$ or $x = 230^{\circ}.$
Therefore, the solution set of the trigonometric equation is $\{50^{\circ}, 170^{\circ}, 230^{\circ}, 350^{\circ}\}.$
(Answer C)

Use the double-angle identity.
$\boxed{\cos 2x = 1 -2 \sin^2 x}$
Therefore, we can write
$\begin{aligned} \cos 2x + \sin x -1 & = 0 \\ (1 -2 \sin^2 x) + \sin x -1 & = 0 \\ -2 \sin^2 x + \sin x & = 0 \\ \sin x(-2 \sin x + 1) & = 0. \end{aligned}$
The last equation shows that
$\sin x = 0 \implies x = 180^{\circ} \lor 360^{\circ}.$
Note that $x \neq 0^{\circ}$, even though $\sin 0^{\circ} = 0$, because it lies outside the interval for $x.$
$\begin{aligned}-2 \sin x + 1 &= 0 \\ \sin x & = \dfrac{1}{2} \\ x & = 30^{\circ} \lor x = 150^{\circ} \end{aligned}$
Therefore, the values of $x$ satisfying the trigonometric equation in set notation are $\boxed{\{30^{\circ}, 150^{\circ}, 180^{\circ}, 360^{\circ}\}}$
(Answer A)

Using the general form of the sine double-angle formula, namely
$\boxed{\sin 2ax = 2 \sin ax \cos ax}$
we obtain
$\begin{aligned} \sin 4x -\cos 2x & = 0 \\ (2 \sin 2x \cos 2x) -\cos 2x & = 0 \\ \cos 2x(2 \sin 2x -1) & = 0. \end{aligned}$
The last equation shows that
$\cos 2x = 0 \implies x = 45^{\circ} \lor 135^{\circ}$
or
$\begin{aligned} 2 \sin 2x -1 & = 0 \\ \sin 2x & = \dfrac{1}{2} \\ x & = 15^{\circ} \lor x = 75^{\circ}. \end{aligned}$
Therefore, the solution set of the trigonometric equation is $\boxed{\{15^{\circ}, 45^{\circ}, 75^{\circ}, 135^{\circ}\}}.$
(Answer C)

Use the identity:
$\boxed{\cos 2x = 2 \cos^2 x -1}$
Thus, we can write
$\begin{aligned} \cos 2x -5 \cos x – 2 & = 0 \\ (2 \cos^2 x -1) -5 \cos x -2 & = 0 \\ 2 \cos^2 x -5 \cos x -3 & = 0 \\ (2 \cos x + 1)(\cos x -3) & = 0. \end{aligned}$
We obtain $2 \cos x + 1 = 0$ (which is equivalent to $\cos x = -\dfrac{1}{2}$) or $\cos x -3 = 0$ (which is equivalent to $\cos x = 3$, but has no solution because the maximum value of cosine is $1$).
Consider the equation $\cos x = -\dfrac{1}{2}$, which can be written as
$\cos x = \cos 120^{\circ}.$
Possibility 1:
$x = 120^{\circ} + k \cdot 360^{\circ}$
For $k=0$, we obtain $x = 120^{\circ} < \pi~~(\text{X})$
For $k=1$, we obtain $x = 480^{\circ}~~(\text{X})$
Possibility 2:
$x = -120^{\circ} + k \cdot 360^{\circ}$
For $k=0$, we obtain $x = -120^{\circ} < \pi~~(\text{X})$
For $k=1$, we obtain $x = 240^{\circ}~~(\checkmark)$
For $k=2$, we obtain $x = 600^{\circ}~~(\text{X})$
Therefore, the only value of $x$ that satisfies the trigonometric equation is $x = 240^{\circ}$ so that $\boxed{\tan x = \tan 240^{\circ} = \sqrt{3}}.$
(Answer A)

Use the identity:
$\boxed{\cos 4x = 2 \cos^2 2x -1}$
Simplify the equation, then rewrite the variable form in terms of $\cos 2x$.
$$\begin{aligned} \sqrt{3-3 \cos^2 2x} -\cos 4x & = 2 \\ \sqrt{3-3 \cos^2 2x} & = 2 + \cos 4x \\ \sqrt{3-3 \cos^2 2x} & = 2 + (2 \cos^2 2x – 1) \\ \sqrt{3-3 \cos^2 2x} & = 1 + 2 \cos^2 2x \\ \text{Square both}~&\text{sides} \\ 3-3 \cos^2 2x & = (1 + 2 \cos^2 2x)^2 \\ \text{Let}~a & = \cos^2 2x \\ 3 -3a & = (1+2a)^2 \\ 3-3a & = 1+4a+4a^2 \\ 4a^2+7a-2 & = 0 \\ (4a-1) (a+2) & = 0 \end{aligned}$$We obtain $a = \dfrac{1}{4}$ or $a = -2.$
Radical condition:
$$\begin{aligned} 3-3 \cos^2 2x \geq 0 \\ -3 \cos^2 2x \geq -3 \\ \cos^2 2x \leq 1 \end{aligned}$$Substitute back $a = \cos^2 2x$. Thus, it can be written as
$\cos^2 2x = \dfrac{1}{4}$ or $\cos^2 2x = -2.$
The second equation has no solution because a square expression cannot produce a negative number.
Now, consider the equation $\cos^2 2x = \dfrac{1}{4}$ (which satisfies the radical condition) equivalent to $\cos 2x = \pm \dfrac{1}{2}.$
For $\cos 2x = \dfrac{1}{2} = \cos 60^{\circ}$ there will be 2 possibilities.
Possibility 1:
$\begin{aligned} 2x & = 60^{\circ} + k \cdot 360^{\circ} \\ x & = 30^{\circ} + k \cdot 180^{\circ} \end{aligned}$
Substitute $k = 0$ and $k=1$ to obtain $x = 30^{\circ}$ and $x = 210^{\circ}$.
Possibility 2:
$\begin{aligned} 2x & = -60^{\circ} + k \cdot 360^{\circ} \\ x & = -30^{\circ} + k \cdot 180^{\circ} \end{aligned}$
Substitute $k = 1$ and $k=2$ to obtain $x = 150^{\circ}$ and $x = 330^{\circ}$.
For $\cos 2x = -\dfrac{1}{2} = \cos 120^{\circ}$ there will also be 2 possibilities.
Possibility 1:
$\begin{aligned} 2x & = 120^{\circ} + k \cdot 360^{\circ} \\ x & = 60^{\circ} + k \cdot 180^{\circ} \end{aligned}$
Substitute $k = 0$ and $k=1$ to obtain $x = 60^{\circ}$ and $x = 240^{\circ}$.
Possibility 2:
$\begin{aligned} 2x & = -120^{\circ} + k \cdot 360^{\circ} \\ x & = -60^{\circ} + k \cdot 180^{\circ} \end{aligned}$
Substitute $k = 1$ and $k=2$ to obtain $x = 120^{\circ}$ and $x = 300^{\circ}$.
Therefore, the solution set of the equation is $$\boxed{\{30^{\circ}, 60^{\circ}, 120^{\circ}, 120^{\circ}, 150^{\circ}, 210^{\circ}, 240^{\circ}, 300^{\circ}, 330^{\circ} \}}$$Hence, the value that is not included in the solution set of the trigonometric equation is $\boxed{180^{\circ}}.$
(Answer D)

The relationship between sine and cosine in Quadrant I is given by:
$\sin (90-x)^{\circ} = \cos x^{\circ}$
Therefore, the equation $\sin (-x+5)^{\circ} = \cos (25-3x)^{\circ}$ can be rewritten as
$\begin{aligned} \cos (90-(-x+5))^{\circ} &= \cos (25-3x)^{\circ} \\ \cos (x+85)^{\circ} & = \cos (25-3x)^{\circ} \end{aligned}$
Next, using the concept of the basic trigonometric equation for cosine, we obtain two possibilities to determine the solutions of the equation.
Possibility 1:
$\begin{aligned} (x+85)^{\circ} & = (25-3x)^{\circ} + k \cdot 360^{\circ} \\ 4x^{\circ} & = -60^{\circ} + k \cdot 360^{\circ} \\ x^{\circ} & = -15^{\circ} + k \cdot 90^{\circ} \end{aligned}$
For certain values of $k$, we obtain the following values of $x$.
$$\begin{array}{|c|c|c|} \hline \text{Value of}~k & \text{Value of}~x & \text{Remark} \\ \hline 0 & -15 & \text{Does Not Satisfy} \\ 1 & 75 & \text{Satisfies} \\ 2 & 165 & \text{Does Not Satisfy} \\ \hline \end{array}$$Note: The obtained value of $x$ is considered valid if it lies in the interval $0^{\circ} \leq x^{\circ} \leq 90^{\circ}$.
Possibility 2:
$\begin{aligned} (x+85)^{\circ} & = -(25-3x)^{\circ} + k \cdot 360^{\circ} \\ -2x^{\circ} & = -110^{\circ} + k \cdot 360^{\circ} \\ x^{\circ} & = 55^{\circ} -k \cdot 180^{\circ} \end{aligned}$
For certain values of $k$, we obtain the following values of $x$.
$$\begin{array}{|c|c|c|} \hline \text{Value of}~k & \text{Value of}~x & \text{Remark} \\ \hline 0 & 55 & \text{Satisfies} \\ 1 & 125 & \text{Does Not Satisfy} \\ \hline \end{array}$$Therefore, the solution set for $x$ is $\{55, 75\}.$
(Answer B)

Notice that secant is the reciprocal form of cosine, so it can be written as
$\begin{aligned} \sec x -2 -15 \cos x & = 0 \\ \dfrac{1}{\cos x} -2 -15 \cos x & = 0 \\ \text{Multiply both sides by}~&\cos x \\ 1 -2 \cos x -15 \cos^2 x & = 0. \end{aligned}$
We obtain a quadratic equation form by letting $a = \cos x$, resulting in $-15a^2 -2a + 1 = 0.$
Since $x_1$ and $x_2$ are the roots (solutions) of the equation, the product of the roots is
$x_1 \cdot x_2 = \dfrac{\text{Constant}} {\text{Coefficient of}~a^2} = -\dfrac{1}{15}.$
Therefore, the value of $\boxed{\dfrac{1}{\cos x_1 \cdot \cos x_2} = -15}.$
(Answer B)

Use the following trigonometric identities/formulas.
$\boxed{\begin{aligned} & \dfrac{\sin ax} {\cos ax} = \tan ax \\ & \dfrac{\cos ax} {\sin ax} = \cot ax \\ & \cot 2x = \dfrac{1- \tan^2 x} {2 \tan x} \\ & \tan (A + B) = \dfrac{\tan A + \tan B} {1 -\tan A \tan B} \end{aligned}}$
Thus, it can be written as
$$\begin{aligned} \dfrac{2 \sin x \cos 2x} {\cos x \sin 2x} -5 \tan x + 5 & = 0 \\ 2 \cdot \dfrac{\sin x} {\cos x} \cdot \dfrac{\cos 2x} {\sin 2x} -5 \tan x + 5 & = 0 \\ 2 \tan x \cot 2x -5 \tan x + 5 & = 0 \\ \cancel{2 \tan x} \cdot \dfrac{1- \tan^2 x} {\cancel{2 \tan x}} -5 \tan x + 5 & = 0 \\ 1 -\tan^2 x -5 \tan x + 5 & = 0 \\ \tan^2 x + 5 \tan x -6 & = 0 \\ (\tan x + 6)(\tan x -1) & = 0 \end{aligned}$$Next, we obtain $\tan x_1 = -6$ or $\tan x_2 = 1.$
Therefore,
$\begin{aligned} \tan (x_1 + x_2) & = \dfrac{\tan x_1 + \tan x_2} {1 -\tan x_1 \tan x_2} \\ & = \dfrac{-6 + 1}{1 -(-6)(1)} \\ & = -\dfrac{5}{7}. \end{aligned}$
Therefore, the value of $\boxed{\tan (x_1 + x_2) = – \dfrac{5}{7}}.$
(Answer A)

Use the following trigonometric identities/formulas.
$\boxed{\begin{aligned} & \dfrac{\sin ax} {\cos ax} = \tan ax \\ & \sec ax = \dfrac{1}{\cos ax} \end{aligned}}$
Thus, it can be written as
$$\begin{aligned} -2 \tan x \sec x -2 \tan x + 5 \sin x & = 0 \\ -2 \tan x(\sec x + 1) + 5 \sin x & = 0 \\ 5 \sin x & = 2 \tan x (\sec x + 1) \\ 5~\cancel{\sin x} & = \dfrac{2~\cancel{\sin x}} {\cos x} \left(\dfrac{1}{\cos x} + 1\right) \\ 5 & = \dfrac{2}{\cos^2 x} + \dfrac{2}{\cos x} \\ \text{Multiply both}~&\text{sides by}~\cos^2 x \\ 5 \cos^2 x & = 2 + 2 \cos x \\ 5 \cos^2 x -2 \cos x -2 & = 0. \end{aligned}$$Using the quadratic formula, we obtain the roots
$$\begin{aligned} \cos x & = \dfrac{-(-2) \pm \sqrt{(-2)^2-4(5)(-2)}}{2(5)} \\ & = \dfrac{2 \pm \sqrt{44}} {10} \end{aligned}$$Next, we check whether both solutions satisfy the interval $0 < x < \pi$.
For $0 < x < \pi$, the possible interval values of $\cos x$ are $-1 < \cos x < 1$ (observe again the graph of the cosine function). The form $-1 < \cos x < 1$ is equivalent to $|\cos x| < 1$
Now, observe that
$$|\cos x| = \left|\dfrac{2 + \sqrt{44}} {10}\right| < \left|\dfrac{2 + \sqrt{49}} {10}\right| = \dfrac{9}{10} < 1$$(Satisfied)
$$|\cos x| = \left|\dfrac{2 – \sqrt{44}} {10}\right| < \left|\dfrac{2 – \sqrt{49}} {10}\right| = \dfrac{5}{10} < 1$$(Satisfied)
Therefore, there are $\boxed{2}$ solutions satisfying the trigonometric equation above (Answer C)

Use the following trigonometric identities/formulas.
$\boxed{\begin{aligned} & \dfrac{\sin ax} {\cos ax} = \tan ax \\ & \sec ax = \dfrac{1}{\cos ax} \end{aligned}}$
Thus, we can write
$$\begin{aligned} 2 \sin x + \sec x -2 \tan x -1 & = 0 \\ 2 \sin x + \dfrac{1}{\cos x} – \dfrac{2 \sin x} {\cos x} -1 & = 0 \\ \text{Multiply both sides by}~& \cos x \\ 2 \sin x \cos x + 1 -2 \sin x -\cos x & = 0 \\ 2 \sin x(\cos x -1) -(\cos x -1) & = 0 \\ (2 \sin x -1)(\cos x -1) & = 0. \end{aligned}$$We obtain $\sin x = \dfrac{1}{2}$ or $\cos x = 1.$
For $\sin x_1 = \dfrac{1}{2}$ and $\cos x_2 = 1$, we get
$\boxed{\sin x_1 + \cos x_2 = \dfrac{1}{2} + 1 = \dfrac{3}{2}}.$
(Answer D)

Use the trigonometric identity:
$\boxed{\cot 2x =\dfrac{\cot^2 x -1}{2 \cot x}}$
Therefore, we obtain
$$\begin{aligned} 2(\cot 2x) (\cot x) + \cot x & = 1 \\ \bcancel{2} \left(\dfrac{\cot^2 x -1}{\bcancel{2} \cancel{\cot x}}\right) (\cancel{\cot x}) + \cot x -1 & = 0 \\ \cot^2 x + \cot x -2 & = 0 \\ (\cot x + 2)(\cot x -1) & = 0 \end{aligned}$$Thus, we obtain
$\cot x = -2$ or $\cot x = 1.$
Therefore, $\boxed{(\cot x_1)(\cot x_2) = (-2)(1) = -2}.$
(Answer A)

Use the following trigonometric identities/formulas.
$\boxed{\begin{aligned} & \cot x = \dfrac{\cos x} {\sin x} \\ & \csc x = \dfrac{1}{\sin x} \\ & \sin^2 x + \cos^2 x = 1 \end{aligned}}$
Thus, we obtain
$\begin{aligned} 2 \sin x + 3 \cot x -3 \csc x & = 0 \\ 2 \sin x +3 \cdot \dfrac{\cos x} {\sin x} -\dfrac{3}{\sin x} & = 0 \\ \text{Multiply both sides by}~&\sin x \\ 2 \sin^2 x + 3 \cos x -3 & = 0 \\ 2(1 -\cos^2 x) + 3 \cos x -3 & = 0 \\ -2 \cos^2 x + 3 \cos x -1 & = 0 \\ (2 \cos x -1)(\cos x -1) & = 0. \end{aligned}$
Thus, we obtain
$\cos x = \dfrac{1}{2}$ or $\cos x = 1.$
Notice that for $0 < x < \dfrac{\pi} {2}$, $\cos x = 1$ has no solution. Therefore, consider only $\cos x = \dfrac{1}{2}$.
Recall that cosine is the ratio $\dfrac{\text{adjacent}} {\text{hypotenuse}}$ so the opposite side length is $\sqrt{2^2-1^2} = \sqrt{3}.$
This means, $\sin x = \dfrac{\sqrt{3}} {2}.$
Therefore, the value of $\boxed{\sin x \cdot \cos x = \dfrac{\sqrt{3}} {2} \cdot \dfrac{1}{2} = \dfrac{1}{4}\sqrt{3}}.$
(Answer D)

Use the following trigonometric identities/formulas.
$\boxed{\begin{aligned} & \sec \theta = \dfrac{1}{\cos \theta} \\ & \tan \theta = \dfrac{\sin \theta} {\cos \theta} \end{aligned}}$
Thus, we can write
$$\begin{aligned} \sec \theta \left(\sec \theta \sin^2 \theta + \dfrac23\sqrt3 \sin \theta \right) & = 1 \\ \sec^2 \theta \sin^2 \theta + \dfrac23\sqrt3 \sec \theta \sin \theta & = 1 \\ \dfrac{1}{\cos^2 \theta} \cdot \sin^2 \theta + \dfrac23\sqrt3 \dfrac{1}{\cos \theta} \cdot \sin \theta & = 1 \\ \tan^2 \theta + \dfrac23\sqrt3 \tan \theta-1 & = 0 \\ \color{red}{3} \tan^2 \theta+2\sqrt3 \tan \theta\color{blue}{-3} & = 0. \end{aligned}$$The last equation can be assumed as a quadratic equation with variable $\tan \theta.$ Suppose its roots are $\tan \theta_1$ and $\tan \theta_2,$ then we obtain the product of the roots as
$$\begin{aligned} \tan \theta_1 \cdot \tan \theta_2 & = \dfrac{c}{a} \\ & = \dfrac{\color{blue}{-3}}{\color{red}{3}} = -1. \end{aligned}$$Therefore, the value of $\boxed{\tan \theta_1 \cdot \tan \theta_2 = -1}.$
(Answer A)

Factor the expression on the left-hand side of the equation.
$\begin{aligned} \csc^2 x + 3 \csc x -10 & = 0 \\ (\csc x + 5)(\csc x -2) & = 0 \end{aligned}$
Thus, we obtain
$\csc x = -5$ or $\csc x = 2.$
Since cosecant is the reciprocal of sine, we obtain
$\sin x = -\dfrac{1}{5}$ or $\sin x = \dfrac{1}{2}.$
For $\sin x_1 = -\dfrac{1}{5}$ and $\sin x_2 = \dfrac{1}{2}$, we obtain
$\begin{aligned} \dfrac{\sin x_1 + \sin x_2}{\sin x_1 \cdot \sin x_2} & = \dfrac{-\dfrac{1}{5} + \dfrac{1}{2}} {-\dfrac{1}{5} \cdot \dfrac{1}{2}} \\ & = \dfrac{-\dfrac{3}{\cancel{10}}} {\dfrac{1}{\cancel{10}}} = -3. \end{aligned}$
Therefore, the value of $\boxed{\dfrac{\sin x_1 + \sin x_2}{\sin x_1 \cdot \sin x_2} = -3}.$
(Answer C)

Notice that the equation $\cot^2 x -6 \cot x = 1$ is equivalent to the equation $\cot^2 x -6 \cot x -1 = 0$
Since the equation can be assumed as a quadratic equation with variable $\cot x$, the roots can be determined using the quadratic formula, namely
$$\begin{aligned} \cot x & = \dfrac{-(-6) \pm \sqrt{(-6)^2-4(1)(-1)}}{2(1)} \\ & = \dfrac{6 \pm \sqrt{40}} {2} \\ & = 3 \pm \sqrt{10}. \end{aligned}$$Cotangent is the ratio of the adjacent side to the opposite side (cot = adjacent/opposite), so to determine the sine value we can use a right triangle approach.
We obtain the hypotenuse lengths as
$\begin{aligned} h_1 & = \sqrt{1^2 + (3 + \sqrt{10})^2} \\ & = \sqrt{1 + (9 + 6\sqrt{10} + 10)} \\ & = \sqrt{20 + 6\sqrt{10}} \end{aligned}$
or
$\begin{aligned} h_2 & = \sqrt{1^2 + (3 -\sqrt{10})^2} \\ & = \sqrt{1 + (9- 6\sqrt{10} + 10)} \\ & = \sqrt{20 -6\sqrt{10}}. \end{aligned}$
Thus,
$\sin x = \dfrac{1}{\sqrt{20 -6\sqrt{10}}}$ or $\sin x = \sqrt{20 + 6\sqrt{10}}.$
For $\sin x_1 = \dfrac{1}{\sqrt{20 -6\sqrt{10}}}$ and $\sin x_2 = \sqrt{20 + 6\sqrt{10}}$, we obtain
$$\begin{aligned} |\sin x_1 \cdot \sin x_2| & = \left|\dfrac{1}{\sqrt{20 +6\sqrt{10}}} \cdot \dfrac{1}{\sqrt{20 -6\sqrt{10}}}\right| \\ & = \left|\dfrac{1}{400 – (36 \cdot 10)}\right| \\ & = \dfrac{1}{\sqrt{40}} = \dfrac{1}{2\sqrt{10}}. \end{aligned}$$Therefore, the value of $\boxed{|\sin x_1 \cdot \sin x_2| = \dfrac{1}{2\sqrt{10}}}.$
(Answer B)

Use the following trigonometric identities/formulas.
$$\boxed{\begin{aligned} & \tan x = \dfrac{\sin x} {\cos x} \\ & \cot x = \dfrac{\cos x} {\sin x} \\ & \sin 2x = 2 \sin x \cos x \\ & \cos 2x = \cos^2 x -\sin^2 x = 1 -2 \sin^2 x \\ & \sin^2 ax + \cos^2 ax = 1 \end{aligned}}$$Thus, we can write
$$\begin{aligned} 2 \cot x -2 \tan x -4 \sin x \cos x & = 0 \\ 2 \cdot \dfrac{\cos x} {\sin x} -2 \cdot \dfrac{\sin x} {\cos x} & = 4 \sin x \cos x \\ \dfrac{2 \cos^2 x -2 \sin^2 x} {\sin x \cos x} & = 4 \sin x \cos x \\ 2(\cos^2 x – \sin^2 x) & = 4 \sin^2 x \cos^2 x \\ 2 \cos 2x & = \sin^2 2x \\ 2 \cos 2x & = 1 -\cos^2 2x \\ \cos^2 2x + 2 \cos 2x -1 & = 0 \end{aligned}$$The last equation can be assumed as a quadratic equation with variable $\cos 2x$. By using the sum of roots formula and trigonometric identities, we obtain
$$\begin{aligned} \cos 2x_1 + \cos 2x_2 & = -2 \\ (1 -2 \sin^2 x_1) + (1 -2 \sin^2 x_2) & = -2 \\ 2 \sin^2 x_1 + 2 \sin^2 x_2 & = 4 \\ \sin^2 x_1 + \sin^2 x_2 & = 2 \end{aligned}$$Therefore, the value of $\boxed{\sin^2 x_1 + \sin^2 x_2 = 2}.$
(Answer D)

Observe that
$\begin{aligned} \tan x + \cot x & = \dfrac{\sin x}{\cos x} + \dfrac{\cos x}{\sin x} \\ & = \dfrac{\color{red}{\sin^2 x + \cos^2 x}}{\sin x \cos x} \\ & = \dfrac{\color{red}{1}}{\sin x \cos x} \\ & = \dfrac{2}{\color{blue}{2 \sin x \cos x}} \\ & = \dfrac{2}{\color{blue}{\sin 2x}} \end{aligned}$
Therefore, the equation $\sin x + \cos x + \tan x + \cot x = \dfrac{2}{\sin 2x}$ is equivalent to
$\begin{aligned} \sin x + \cos x + \cancel{\dfrac{2}{\sin 2x}} & = \cancel{\dfrac{2}{\sin 2x}} \\ \sin x + \cos x & = 0 \\ \text{Divide both sides by}~& \cos x \\ \tan x + 1 & = 0 \\ \tan x & = -1 \\ \tan x & = \tan 135^{\circ}. \end{aligned}$
We obtain $x = 135^{\circ} + k \cdot 180^{\circ}$ for integer values of $k$ or mathematically written as $k \in \mathbb{Z}.$
(Answer D)

Using the angle sum identity for trigonometric functions, we obtain
$$\begin{aligned} \sin \left(x + \dfrac{\pi}{6}\right) & = \sqrt{3} \sin x \\ \sin (x + 30^{\circ}) & = \sqrt{3} \sin x \\ \sin x \cos 30^{\circ} + \cos x \sin 30^{\circ} & = \sqrt{3} \sin x \\ \sin x \left(\dfrac12\sqrt3\right) + \cos x\left(\dfrac12\right) & = \sqrt{3} \sin x \\ \cancel{\dfrac12} \cos x & = \cancel{\dfrac12}\sqrt{3} \sin x \\ 1 & = \sqrt{3} \cdot \dfrac{\sin x}{\cos x} \\ \dfrac{1}{\sqrt3} = \dfrac13\sqrt3 & = \tan x \end{aligned}$$Thus, it can be determined that the values of $x$ satisfying the equation above are $\boxed{\dfrac{\pi}{6}}$ and $\boxed{\dfrac{7\pi}{6}}.$
(Answer C)

The following trigonometric identities are used to solve the problem above.
$$\boxed{\begin{aligned} \cot x & = \dfrac{1}{\tan x} \\ \tan 2x & = \dfrac{2 \tan x}{1- \tan^2 x} \\\sin^2 x + \cos^2 x & = 1 \\ \tan x & = \dfrac{\sin x}{\cos x} \\ \sec x & = \dfrac{1}{\cos x} \\ 1+\tan^2 x & = \sec^2 x \\ \sin 2x & = 2 \sin x \cos x \end{aligned}}$$We will obtain
$$\begin{aligned} \dfrac{4 \sin^2 x}{\tan x + \cot x} + \dfrac{2 \cos^2 x}{\cot x-\cot 2x} & = \dfrac{\cot x+\tan x}{\cot x-\tan x} \\ \dfrac{4 \sin^2 x}{\tan x + \dfrac{1}{\tan x}} + \dfrac{2 \cos^2 x}{\dfrac{1}{\tan x}-\dfrac{1}{\tan 2x}} & = \dfrac{\dfrac{1}{\tan x}+\tan x}{\dfrac{1}{\tan x}-\tan x} \\ \dfrac{4 \sin^2 x \tan x}{\tan^2 x + 1} + \dfrac{2 \cos^2 x}{\dfrac{1}{\tan x}-\dfrac{1-\tan^2 x}{2 \tan x}} & = \dfrac{1+\tan^2 x}{1-\tan^2 x} \\ \dfrac{4 \sin^2 x \tan x + (2 \cos^2 x)(2 \tan x)}{1+\tan^2 x} & = \dfrac{1+\tan^2 x}{1-\tan^2 x} \\ \dfrac{4 \tan x(\sin^2 x + \cos^2 x)}{\sec^2 x} & = \dfrac{\sec^2 x}{1-\tan^2 x} \\ \dfrac{4 \tan x(1)}{\dfrac{1}{\cos^2 x}} & = \dfrac{\dfrac{1}{\cos^2 x}}{1-\dfrac{\sin^2 x}{\cos^2 x}} \\ 4 \cdot \dfrac{\sin x}{\cos x} \cdot \cos^2 x & = \dfrac{1}{\cos 2x} \\ 4 \sin x \cos x & = \dfrac{1}{\cos 2x} \\ 2 \sin 2x & = \dfrac{1}{\cos 2x} \\ 2 \sin 2x \cos 2x & = 1 \\ \sin 4x & = 1 \\ \sin 4x & = \sin \dfrac{\pi}{2} \\ 4x & = \dfrac{\pi}{2} + k \cdot 2\pi \\ x & = \dfrac{\pi}{8} + k \cdot \dfrac{\pi}{2}. \end{aligned}$$Therefore, the solution of the trigonometric equation is $\boxed{x = \dfrac{\pi}{8} + k \cdot \dfrac{\pi}{2}}.$
(Answer E)

Given $\cos x = \cos \dfrac{2\pi} {5}$. Thus, we obtain the following possibilities.
Possibility 1:
$x = \dfrac{2\pi} {5} + k \cdot 2\pi$
For $k = 0$, we obtain $x = \dfrac{2\pi} {5}~~(\checkmark)$
For $k=1$, we obtain $x > 2\pi~~(\text{X})$
Possibility 2:
$x = -\dfrac{2\pi} {5} + k \cdot 2\pi$
For $k = 0$, we obtain $x = – \dfrac{2\pi} {5}~~(\text{X})$
For $k=1$, we obtain $x = -\dfrac{2\pi} {5} + 2\pi = \dfrac{8\pi} {5}~~(\checkmark)$
For $k=2$, we obtain $x > 2\pi~~(\text{X})$
Therefore, the solution set of the trigonometric equation is $\left\{\dfrac{2\pi} {5}, \dfrac{8\pi} {5}\right\}.$

Expand the expression $(a+1)^2$, then simplify the equation.
$\begin{aligned} 1 + a \cos x & = (a + 1)^2 \\ \cancel{1} + a \cos x & = a^2 + 2a + \cancel{1} \\ \cancel{a} \cos x & = \cancel{a} (a + 2) \\ \cos x & = a + 2 \end{aligned}$
The interval of cosine values is $-1 \leq \cos x \leq 1$, with integer values $\{-1, 0, 1\}.$
For $\cos x = -1$, it means that $a = -3$.
For $\cos x = 0$, it means that $a = -2$.
For $\cos x = 1$, it means that $a = -1$.
Therefore, the integer values of $a$ such that the equation has a solution are $\boxed{\{-3, -2, -1\}}.$

Given $81^{\sin^2 \theta} + 81^{\cos^2 \theta} = 30.$
Suppose $81^{\sin^2 \theta} = a$ and $81^{\cos^2 \theta} = b$, then we obtain $a + b = 30$ and
$$\begin{aligned} ^{81} \log a + \! ^{81} \log b & = \sin^2 \theta + \cos^2 \theta \\ ^{81} \log ab & = 1 \\ ab & = 81. \end{aligned}$$The pairs of values $(a, b)$ that satisfy are $(a, b) = (27, 3)$ or $(a, b) = (3, 27).$

First Case:
For $a = 27$, we obtain
$$\begin{aligned} 81^{\sin^2 \theta} & = 27 \\ 3^{4 \sin^2 \theta} & = 3^3 \\ 4 \sin^2 \theta & = 3 \\ \sin^2 \theta & = \dfrac34 \\ \sin \theta & = \pm \dfrac12\sqrt3 \\ \theta & = 60^\circ, 120^\circ. \end{aligned}$$For $b = 3$, we obtain
$$\begin{aligned} 81^{\cos^2 \theta} & = 3 \\ 3^{4 \cos^2 \theta} & = 3^1 \\ 4 \cos^2 \theta & = 1 \\ \cos^2 \theta & = \dfrac14 \\ \cos \theta & = \pm \dfrac12 \\ \theta & = 60^\circ, 120^\circ. \end{aligned}$$Therefore, the values of $\theta$ satisfying both are $60^\circ$ and $120^\circ$.

Second Case:
For $a = 3$, we obtain
$$\begin{aligned} 81^{\sin^2 \theta} & = 3 \\ 3^{4 \sin^2 \theta} & = 3^1 \\ 4 \sin^2 \theta & = 1 \\ \sin^2 \theta & = \dfrac14 \\ \sin \theta & = \pm \dfrac12 \\ \theta & = 30^\circ, 150^\circ. \end{aligned}$$For $b = 27$, we obtain
$$\begin{aligned} 81^{\cos^2 \theta} & = 27 \\ 3^{4 \cos^2 \theta} & = 3^3 \\ 4 \cos^2 \theta & = 3 \\ \cos^2 \theta & = \dfrac34 \\ \cos \theta & = \pm \dfrac12\sqrt3 \\ \theta & = 30^\circ, 150^\circ. \end{aligned}$$Therefore, the values of $\theta$ satisfying both are $30^\circ$ and $150^\circ.$

Therefore, the values of $\theta$ satisfying the equation $81^{\sin^2 \theta} + 81^{\cos^2 \theta} = 30$ in the interval $0^\circ \le \theta \le 180^\circ$ are $\boxed{30^\circ, 60^\circ, 120^\circ, 150^\circ}.$

The first step is to multiply both sides of the equation above by $(1 -\cos \alpha)$ so that it becomes
$$\begin{aligned} \color{red}{(1-\cos \alpha)} (1 + \cos \alpha) (1+ \cos 2\alpha) (1 + \cos 4\alpha) & = \dfrac{1}{8} \color{red} {(1- \cos \alpha)} \\ \color{red}{ (1 -\cos^2 \alpha)} (1+ \cos 2\alpha) (1 + \cos 4\alpha) & = \dfrac{1}{8}(1- \cos \alpha) \end{aligned}$$The identity $\cos 2\alpha = 2 \cos^2 \alpha -1$ can be rewritten as $1 -\cos^2 \alpha = \dfrac{1 -\cos 2\alpha} {2}$ so that the equation above becomes$$\begin{aligned} \color{red}{\left(\dfrac{1-\cos 2\alpha} {\cancel{2}}\right)} (1+ \cos 2\alpha) (1 + \cos 4\alpha) & = \dfrac{1}{\cancelto{4}{8}}(1-\cos \alpha) \\ \color{red} {(1 -\cos^2 2\alpha)} (1+ \cos 4\alpha) & = \dfrac{1}{4}(1- \cos \alpha) \\ \color{red}{\left(\dfrac{1-\cos 4\alpha} {\cancel{2}}\right)} (1+ \cos 4\alpha) & = \dfrac{1}{\cancelto{2}{4}}(1- \cos \alpha) \\ \color{red}{(1 -\cos^2 4\alpha)} & = \dfrac{1}{2}(1 -\cos \alpha) \\ \color{red}{\left(\dfrac{1-\cos 8\alpha} {\cancel{2}}\right)} & = \dfrac{1}{\cancel{2}}(1- \cos \alpha) \\ \cos 8\alpha & = \cos \alpha \end{aligned}$$We obtain a basic trigonometric equation in cosine form.

Solution: $8\alpha = \pm \alpha + k \cdot 360^{\circ}$ with $0^{\circ} < a < 90^{\circ}$.

Possibility 1:
$\begin{aligned} 8\alpha & = \alpha + k \cdot 360^{\circ} \\ 7\alpha & = k \cdot 360^{\circ} \\ \alpha & = \dfrac{k \cdot 360^{\circ}} {7} \end{aligned}$
For $k = 0$, we obtain $\alpha = 0^{\circ}~~(\text{X}) $
For $k = 1$, we obtain $\alpha = \dfrac{1 \times 360^{\circ}} {7}~~(\checkmark)$
For $k = 2$, we obtain $\alpha = \dfrac{2 \times 360^{\circ}} {7} > 90^{\circ}~~(\text{X})$

Possibility 2:
$\begin{aligned} 8\alpha & = -\alpha + k \cdot 360^{\circ} \\ 9\alpha & = k \cdot 360^{\circ} \\ \alpha & = \dfrac{k \cdot 360^{\circ}} {9} = k \cdot 40^{\circ} \end{aligned}$
For $k = 0$, we obtain $\alpha = 0^{\circ}~~(\text{X}) $
For $k = 1$, we obtain $\alpha = 40^{\circ}~~(\checkmark)$
For $k = 2$, we obtain $\alpha = 80^{\circ}~~(\checkmark)$
For $k = 3$, we obtain $\alpha = 120^{\circ} > 90^{\circ}~~(\text{X})$

Therefore, the number of values of $\alpha$ satisfying the equation is $\boxed{3}$ (see the number of check marks).