Solution of trigonometrical equations Questions in English

(A) Given $\tan n\theta = \tan m\theta$.
The general solution for $\tan x = \tan y$ is $x = N\pi + y$,where $N \in \mathbb{Z}$.
Therefore,$n\theta = N\pi + m\theta$.
Rearranging the terms,we get $(n - m)\theta = N\pi$.
Thus,$\theta = \frac{N\pi}{n - m}$.
For different integer values of $N = 1, 2, 3, \dots$,the values of $\theta$ are $\frac{\pi}{n - m}, \frac{2\pi}{n - m}, \frac{3\pi}{n - m}, \dots$.
This sequence has a common difference $d = \frac{\pi}{n - m}$,which implies that the values are in $A.P.$

(D) Given the quadratic equation $8\sec^2 \theta - 6\sec \theta + 1 = 0$.
Let $x = \sec \theta$. The equation becomes $8x^2 - 6x + 1 = 0$.
Factoring the quadratic: $8x^2 - 4x - 2x + 1 = 0 \implies 4x(2x - 1) - 1(2x - 1) = 0$.
This gives $(4x - 1)(2x - 1) = 0$.
So,$x = \frac{1}{4}$ or $x = \frac{1}{2}$.
Thus,$\sec \theta = \frac{1}{4}$ or $\sec \theta = \frac{1}{2}$.
However,for any real $\theta$,the range of $\sec \theta$ is $(-\infty, -1] \cup [1, \infty)$.
Since both $\frac{1}{4}$ and $\frac{1}{2}$ lie in the interval $(-1, 1)$,there is no real value of $\theta$ that satisfies these equations.
Therefore,the number of roots is $0$.

(D) Given expression: $\cos y \cos (\frac{\pi}{2} - x) - \cos (\frac{\pi}{2} - y) \cos x + \sin y \cos (\frac{\pi}{2} - x) + \cos x \sin (\frac{\pi}{2} - y) = 0$
Using $\cos (\frac{\pi}{2} - \theta) = \sin \theta$ and $\sin (\frac{\pi}{2} - \theta) = \cos \theta$,the expression becomes:
$\cos y \sin x - \sin y \cos x + \sin y \sin x + \cos x \cos y = 0$
Rearranging the terms:
$(\sin x \cos y - \cos x \sin y) + (\sin x \sin y + \cos x \cos y) = 0$
Using the identities $\sin(A - B) = \sin A \cos B - \cos A \sin B$ and $\cos(A - B) = \cos A \cos B + \sin A \sin B$:
$\sin(x - y) + \cos(x - y) = 0$
Dividing by $\cos(x - y)$ (assuming $\cos(x - y) \neq 0$):
$\tan(x - y) = -1$
$\tan(x - y) = \tan(-\frac{\pi}{4})$
$x - y = n\pi - \frac{\pi}{4}$
$x = n\pi - \frac{\pi}{4} + y, (n \in I)$

(D) Given $\cos(\alpha - \beta) = 1$ and $-\pi < \alpha, \beta < \pi$.
Since $-\pi < \alpha < \pi$ and $-\pi < \beta < \pi$,we have $-2\pi < \alpha - \beta < 2\pi$.
The equation $\cos(\alpha - \beta) = 1$ implies $\alpha - \beta = 0$,so $\alpha = \beta$.
Substituting $\alpha = \beta$ into the second equation,we get $\cos(\alpha + \alpha) = \cos(2\alpha) = \frac{1}{e}$.
Since $-\pi < \alpha < \pi$,we have $-2\pi < 2\alpha < 2\pi$.
In the interval $(-2\pi, 2\pi)$,the equation $\cos(2\alpha) = \frac{1}{e}$ has solutions where $\cos(x) = \frac{1}{e}$ for $x \in (-2\pi, 2\pi)$.
Since $0 < \frac{1}{e} < 1$,there are two solutions for $2\alpha$ in $(0, 2\pi)$ and two solutions in $(-2\pi, 0)$.
Thus,there are $4$ distinct values for $\alpha$,and since $\alpha = \beta$,there are $4$ ordered pairs $(\alpha, \beta)$.

(B) Given $\sin \theta + \cos \theta = 1$.
Dividing both sides by $\sqrt{1^2 + 1^2} = \sqrt{2}$,we get:
$\frac{1}{\sqrt{2}} \sin \theta + \frac{1}{\sqrt{2}} \cos \theta = \frac{1}{\sqrt{2}}$
$\sin \theta \cos \frac{\pi}{4} + \cos \theta \sin \frac{\pi}{4} = \sin \frac{\pi}{4}$
$\sin \left( \theta + \frac{\pi}{4} \right) = \sin \frac{\pi}{4}$
The general solution for $\sin x = \sin \alpha$ is $x = n\pi + (-1)^n \alpha$.
Therefore,$\theta + \frac{\pi}{4} = n\pi + (-1)^n \frac{\pi}{4}$
$\theta = n\pi + (-1)^n \frac{\pi}{4} - \frac{\pi}{4}$.

(C) Given $\sin^2 \theta = \frac{1}{4}.$
We know that $\sin^2 \theta = \sin^2 \alpha \implies \theta = n\pi \pm \alpha.$
Since $\sin^2 \theta = \frac{1}{4} = \left(\frac{1}{2}\right)^2 = \sin^2 \left(\frac{\pi}{6}\right),$
we have $\alpha = \frac{\pi}{6}.$
Therefore,the general solution is $\theta = n\pi \pm \frac{\pi}{6},$ where $n \in \mathbb{Z}.$

(B) Given $\sec^2 \theta = \frac{4}{3}$.
Taking the reciprocal,we get $\cos^2 \theta = \frac{3}{4}$.
This can be written as $\cos^2 \theta = \left( \frac{\sqrt{3}}{2} \right)^2 = \cos^2 \left( \frac{\pi}{6} \right)$.
The general solution for $\cos^2 \theta = \cos^2 \alpha$ is $\theta = n\pi \pm \alpha$.
Therefore,$\theta = n\pi \pm \frac{\pi}{6}$.

(B) Given equation: $\cot \theta - \tan \theta = 2$
Using $\cot \theta = \frac{1}{\tan \theta}$,we get: $\frac{1}{\tan \theta} - \tan \theta = 2$
$\Rightarrow \frac{1 - \tan^2 \theta}{\tan \theta} = 2$
$\Rightarrow 1 - \tan^2 \theta = 2 \tan \theta$
$\Rightarrow \tan^2 \theta + 2 \tan \theta - 1 = 0$
Alternatively,using $\cot \theta - \tan \theta = \frac{\cos \theta}{\sin \theta} - \frac{\sin \theta}{\cos \theta} = \frac{\cos^2 \theta - \sin^2 \theta}{\sin \theta \cos \theta} = \frac{\cos 2\theta}{\frac{1}{2} \sin 2\theta} = 2 \cot 2\theta$
So,$2 \cot 2\theta = 2 \Rightarrow \cot 2\theta = 1$
$\Rightarrow \tan 2\theta = 1 = \tan \frac{\pi}{4}$
General solution for $\tan x = \tan \alpha$ is $x = n\pi + \alpha$
Therefore,$2\theta = n\pi + \frac{\pi}{4}$
$\Rightarrow \theta = \frac{n\pi}{2} + \frac{\pi}{8}$

(D) Given equation: $\sqrt{3} \cos \theta + \sin \theta = \sqrt{2}$.
Divide both sides by $\sqrt{(\sqrt{3})^2 + 1^2} = 2$:
$\frac{\sqrt{3}}{2} \cos \theta + \frac{1}{2} \sin \theta = \frac{\sqrt{2}}{2}$.
This can be written as $\sin \frac{\pi}{3} \cos \theta + \cos \frac{\pi}{3} \sin \theta = \frac{1}{\sqrt{2}}$.
Using the formula $\sin(A + B) = \sin A \cos B + \cos A \sin B$,we get:
$\sin(\theta + \frac{\pi}{3}) = \sin \frac{\pi}{4}$.
The general solution for $\sin x = \sin \alpha$ is $x = n\pi + (-1)^n \alpha$.
Therefore,$\theta + \frac{\pi}{3} = n\pi + (-1)^n \frac{\pi}{4}$.
$\theta = n\pi + (-1)^n \frac{\pi}{4} - \frac{\pi}{3}$.

(B) Given the equation: $\sin^2 \theta - 2\cos \theta + \frac{1}{4} = 0$
Using the identity $\sin^2 \theta = 1 - \cos^2 \theta$,we get:
$1 - \cos^2 \theta - 2\cos \theta + \frac{1}{4} = 0$
$-\cos^2 \theta - 2\cos \theta + \frac{5}{4} = 0$
Multiplying by $-1$:
$\cos^2 \theta + 2\cos \theta - \frac{5}{4} = 0$
Using the quadratic formula $\cos \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$\cos \theta = \frac{-2 \pm \sqrt{4 - 4(1)(-\frac{5}{4})}}{2} = \frac{-2 \pm \sqrt{4 + 5}}{2} = \frac{-2 \pm 3}{2}$
This gives two possible values:
$1) \cos \theta = \frac{-2 + 3}{2} = \frac{1}{2}$
$2) \cos \theta = \frac{-2 - 3}{2} = -\frac{5}{2}$
Since $|\cos \theta| \le 1$,the value $\cos \theta = -\frac{5}{2}$ is rejected.
Thus,$\cos \theta = \frac{1}{2} = \cos(\frac{\pi}{3})$.
The general solution for $\cos \theta = \cos \alpha$ is $\theta = 2n\pi \pm \alpha$.
Therefore,$\theta = 2n\pi \pm \frac{\pi}{3}$.

(C) Given: $\sqrt{2} \sec \theta + \tan \theta = 1$
$\Rightarrow \frac{\sqrt{2}}{\cos \theta} + \frac{\sin \theta}{\cos \theta} = 1$
$\Rightarrow \sqrt{2} + \sin \theta = \cos \theta$
$\Rightarrow \sin \theta - \cos \theta = -\sqrt{2}$
Dividing both sides by $\sqrt{2}$,we get:
$\frac{1}{\sqrt{2}} \sin \theta - \frac{1}{\sqrt{2}} \cos \theta = -1$
Multiplying by $-1$:
$\frac{1}{\sqrt{2}} \cos \theta - \frac{1}{\sqrt{2}} \sin \theta = 1$
$\Rightarrow \cos \theta \cos \frac{\pi}{4} - \sin \theta \sin \frac{\pi}{4} = 1$
$\Rightarrow \cos \left( \theta + \frac{\pi}{4} \right) = 1$
Since $\cos \alpha = 1 \Rightarrow \alpha = 2n\pi$,we have:
$\theta + \frac{\pi}{4} = 2n\pi$
$\Rightarrow \theta = 2n\pi - \frac{\pi}{4}$.

(C) Given the equation: $2\tan^2 \theta = \sec^2 \theta$
Using the identity $\sec^2 \theta = 1 + \tan^2 \theta$,we get:
$2\tan^2 \theta = 1 + \tan^2 \theta$
Subtracting $\tan^2 \theta$ from both sides:
$\tan^2 \theta = 1$
This can be written as:
$\tan^2 \theta = \tan^2 \left(\frac{\pi}{4}\right)$
For $\tan^2 \theta = \tan^2 \alpha$,the general solution is $\theta = n\pi \pm \alpha$.
Therefore,$\theta = n\pi \pm \frac{\pi}{4}$.

(B) Given the equation: $2\sin \theta + \tan \theta = 0$
We can write $\tan \theta$ as $\frac{\sin \theta}{\cos \theta}$:
$2\sin \theta + \frac{\sin \theta}{\cos \theta} = 0$
Factor out $\sin \theta$:
$\sin \theta \left( 2 + \frac{1}{\cos \theta} \right) = 0$
This gives two cases:
Case $1$: $\sin \theta = 0 \Rightarrow \theta = n\pi$,where $n \in \mathbb{Z}$.
Case $2$: $2 + \frac{1}{\cos \theta} = 0$ $\Rightarrow \frac{1}{\cos \theta} = -2$ $\Rightarrow \cos \theta = -\frac{1}{2}$.
Since $\cos \theta = -\frac{1}{2} = \cos \left( \frac{2\pi}{3} \right)$,the general solution is $\theta = 2n\pi \pm \frac{2\pi}{3}$,where $n \in \mathbb{Z}$.
Combining both cases,the general values of $\theta$ are $n\pi$ and $2n\pi \pm \frac{2\pi}{3}$.

(B) Given equation: $\sqrt{3} \tan 2\theta + \sqrt{3} \tan 3\theta + \tan 2\theta \tan 3\theta = 1$
Rearranging the terms: $\sqrt{3}(\tan 2\theta + \tan 3\theta) = 1 - \tan 2\theta \tan 3\theta$
Dividing both sides by $\sqrt{3}(1 - \tan 2\theta \tan 3\theta)$,we get:
$\frac{\tan 2\theta + \tan 3\theta}{1 - \tan 2\theta \tan 3\theta} = \frac{1}{\sqrt{3}}$
Using the formula $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we have:
$\tan(2\theta + 3\theta) = \tan\left(\frac{\pi}{6}\right)$
$\tan 5\theta = \tan\left(\frac{\pi}{6}\right)$
The general solution for $\tan x = \tan \alpha$ is $x = n\pi + \alpha$.
Therefore,$5\theta = n\pi + \frac{\pi}{6}$
$\theta = \frac{n\pi}{5} + \frac{\pi}{30} = \left(n + \frac{1}{6}\right)\frac{\pi}{5}$.

(A) Given the equation: $\tan 2\theta \tan \theta = 1$
This can be rewritten as: $\tan 2\theta = \frac{1}{\tan \theta} = \cot \theta$
Using the identity $\cot \theta = \tan \left( \frac{\pi}{2} - \theta \right)$,we get:
$\tan 2\theta = \tan \left( \frac{\pi}{2} - \theta \right)$
The general solution for $\tan x = \tan \alpha$ is $x = n\pi + \alpha$,where $n \in \mathbb{Z}$.
Therefore: $2\theta = n\pi + \left( \frac{\pi}{2} - \theta \right)$
Adding $\theta$ to both sides: $3\theta = n\pi + \frac{\pi}{2}$
Dividing by $3$: $\theta = \frac{n\pi}{3} + \frac{\pi}{6} = \left( n + \frac{1}{2} \right) \frac{\pi}{3}$.

(C) Given equation: $1 + \cot \theta = \text{cosec} \theta$
Rewrite in terms of $\sin \theta$ and $\cos \theta$: $1 + \frac{\cos \theta}{\sin \theta} = \frac{1}{\sin \theta}$
Multiply by $\sin \theta$ (where $\sin \theta \neq 0$): $\sin \theta + \cos \theta = 1$
Divide by $\sqrt{2}$: $\frac{1}{\sqrt{2}} \sin \theta + \frac{1}{\sqrt{2}} \cos \theta = \frac{1}{\sqrt{2}}$
$\cos \frac{\pi}{4} \sin \theta + \sin \frac{\pi}{4} \cos \theta = \frac{1}{\sqrt{2}}$
$\sin(\theta + \frac{\pi}{4}) = \sin \frac{\pi}{4}$
General solution for $\sin x = \sin \alpha$ is $x = n\pi + (-1)^n \alpha$:
Case $1$: If $n$ is even,let $n = 2k$: $\theta + \frac{\pi}{4} = 2k\pi + \frac{\pi}{4} \Rightarrow \theta = 2k\pi$
Case $2$: If $n$ is odd,let $n = 2k+1$: $\theta + \frac{\pi}{4} = (2k+1)\pi - \frac{\pi}{4}$ $\Rightarrow \theta = 2k\pi + \pi - \frac{\pi}{2} = 2k\pi + \frac{\pi}{2}$
Check domain: If $\theta = 2k\pi$,then $\cot \theta$ and $\text{cosec} \theta$ are undefined. Thus,$\theta = 2n\pi$ is rejected.
Therefore,the general solution is $\theta = 2n\pi + \frac{\pi}{2}$.

(D) Given the equation: $\frac{1 - \cos 2\theta}{1 + \cos 2\theta} = 3$
Using the trigonometric identities $\cos 2\theta = 1 - 2\sin^2 \theta$ and $\cos 2\theta = 2\cos^2 \theta - 1$,we get:
$\frac{1 - (1 - 2\sin^2 \theta)}{1 + (2\cos^2 \theta - 1)} = 3$
$\frac{2\sin^2 \theta}{2\cos^2 \theta} = 3$
$\tan^2 \theta = 3$
$\tan^2 \theta = (\sqrt{3})^2 = \tan^2(\frac{\pi}{3})$
The general solution for $\tan^2 \theta = \tan^2 \alpha$ is $\theta = n\pi \pm \alpha$.
Therefore,$\theta = n\pi \pm \frac{\pi}{3}$.

(C) Given $3(\sec^2 \theta + \tan^2 \theta) = 5$,we have $\sec^2 \theta + \tan^2 \theta = \frac{5}{3}$.
We know the identity $\sec^2 \theta - \tan^2 \theta = 1$.
Subtracting the identity from the given equation: $(\sec^2 \theta + \tan^2 \theta) - (\sec^2 \theta - \tan^2 \theta) = \frac{5}{3} - 1$.
$2 \tan^2 \theta = \frac{2}{3} \Rightarrow \tan^2 \theta = \frac{1}{3}$.
Since $\tan^2 \theta = \frac{1}{3} = \tan^2(\frac{\pi}{6})$,the general solution for $\tan^2 \theta = \tan^2 \alpha$ is $\theta = n\pi \pm \alpha$.
Therefore,$\theta = n\pi \pm \frac{\pi}{6}$.

(C) Given equation: $\cos 7\theta = \cos \theta - \sin 4\theta$
Rearranging the terms: $\sin 4\theta = \cos \theta - \cos 7\theta$
Using the formula $\cos C - \cos D = 2 \sin \frac{C+D}{2} \sin \frac{D-C}{2}$:
$\sin 4\theta = 2 \sin \frac{\theta + 7\theta}{2} \sin \frac{7\theta - \theta}{2}$
$\sin 4\theta = 2 \sin 4\theta \sin 3\theta$
$\sin 4\theta (1 - 2 \sin 3\theta) = 0$
Case $1$: $\sin 4\theta = 0$ $\Rightarrow 4\theta = n\pi$ $\Rightarrow \theta = \frac{n\pi}{4}$
Case $2$: $1 - 2 \sin 3\theta = 0 \Rightarrow \sin 3\theta = \frac{1}{2} = \sin \frac{\pi}{6}$
Using the general solution $\sin x = \sin \alpha \Rightarrow x = n\pi + (-1)^n \alpha$:
$3\theta = n\pi + (-1)^n \frac{\pi}{6} \Rightarrow \theta = \frac{n\pi}{3} + (-1)^n \frac{\pi}{18}$
Thus,the general values are $\theta = \frac{n\pi}{4}, \frac{n\pi}{3} + (-1)^n \frac{\pi}{18}$.

(A) Given: $\frac{1 - \tan^2 \theta}{\sec^2 \theta} = \frac{1}{2}$
Since $\frac{1}{\sec^2 \theta} = \cos^2 \theta$ and $\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}$,we have:
$\cos^2 \theta (1 - \frac{\sin^2 \theta}{\cos^2 \theta}) = \frac{1}{2}$
$\cos^2 \theta - \sin^2 \theta = \frac{1}{2}$
Using the identity $\cos 2\theta = \cos^2 \theta - \sin^2 \theta$,we get:
$\cos 2\theta = \frac{1}{2} = \cos(\frac{\pi}{3})$
The general solution for $\cos x = \cos \alpha$ is $x = 2n\pi \pm \alpha$.
Therefore,$2\theta = 2n\pi \pm \frac{\pi}{3}$
Dividing by $2$,we get $\theta = n\pi \pm \frac{\pi}{6}$.

(D) Given the equation: $\cos \theta + \sec \theta = \frac{5}{2}$.
Since $\sec \theta = \frac{1}{\cos \theta}$,we have $\cos \theta + \frac{1}{\cos \theta} = \frac{5}{2}$.
Multiplying by $2 \cos \theta$,we get $2 \cos^2 \theta + 2 = 5 \cos \theta$,which simplifies to $2 \cos^2 \theta - 5 \cos \theta + 2 = 0$.
Factoring the quadratic equation: $(2 \cos \theta - 1)(\cos \theta - 2) = 0$.
This gives $\cos \theta = \frac{1}{2}$ or $\cos \theta = 2$.
Since the range of $\cos \theta$ is $[-1, 1]$,we reject $\cos \theta = 2$.
Thus,$\cos \theta = \frac{1}{2} = \cos \left( \frac{\pi}{3} \right)$.
The general solution for $\cos \theta = \cos \alpha$ is $\theta = 2n\pi \pm \alpha$.
Therefore,$\theta = 2n\pi \pm \frac{\pi}{3}$.

(C) Given: $\cot \theta + \tan \theta = 2 \csc \theta$
Converting to $\sin$ and $\cos$: $\frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta} = \frac{2}{\sin \theta}$
Simplify the left side: $\frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta} = \frac{2}{\sin \theta}$
Since $\cos^2 \theta + \sin^2 \theta = 1$: $\frac{1}{\sin \theta \cos \theta} = \frac{2}{\sin \theta}$
Assuming $\sin \theta \neq 0$,we can cancel $\sin \theta$ from both sides: $\frac{1}{\cos \theta} = 2$
$\cos \theta = \frac{1}{2}$
The general solution for $\cos \theta = \cos \alpha$ is $\theta = 2n\pi \pm \alpha$.
Since $\cos \theta = \frac{1}{2} = \cos(\frac{\pi}{3})$,the general value is $\theta = 2n\pi \pm \frac{\pi}{3}$.

(A) Given the equation: $\tan^2 \theta - (1 + \sqrt{3}) \tan \theta + \sqrt{3} = 0$
Factor the quadratic equation:
$\tan^2 \theta - \tan \theta - \sqrt{3} \tan \theta + \sqrt{3} = 0$
$\tan \theta (\tan \theta - 1) - \sqrt{3} (\tan \theta - 1) = 0$
$(\tan \theta - \sqrt{3})(\tan \theta - 1) = 0$
This gives two cases:
$1) \tan \theta = \sqrt{3}$ $\Rightarrow \tan \theta = \tan(\frac{\pi}{3})$ $\Rightarrow \theta = n\pi + \frac{\pi}{3}$
$2) \tan \theta = 1$ $\Rightarrow \tan \theta = \tan(\frac{\pi}{4})$ $\Rightarrow \theta = n\pi + \frac{\pi}{4}$
Thus,the general solution is $\theta = n\pi + \frac{\pi}{3}, n\pi + \frac{\pi}{4}$.

(A) Given the equations $\sin \theta = \sin \alpha$ and $\cos \theta = \cos \alpha$.
From $\sin \theta = \sin \alpha$,the general solution is $\theta = n\pi + (-1)^n \alpha$,where $n \in \mathbb{Z}$.
From $\cos \theta = \cos \alpha$,the general solution is $\theta = 2m\pi \pm \alpha$,where $m \in \mathbb{Z}$.
For $\theta$ to satisfy both equations simultaneously,the angle must be coterminal with $\alpha$.
This implies $\theta = 2n\pi + \alpha$ for any integer $n$.

(A) Given equation: $4\sin^2 \theta + 2(\sqrt{3} + 1)\cos \theta = 4 + \sqrt{3}$
Substitute $\sin^2 \theta = 1 - \cos^2 \theta$:
$4(1 - \cos^2 \theta) + 2(\sqrt{3} + 1)\cos \theta = 4 + \sqrt{3}$
$4 - 4\cos^2 \theta + 2(\sqrt{3} + 1)\cos \theta = 4 + \sqrt{3}$
$4\cos^2 \theta - 2(\sqrt{3} + 1)\cos \theta + \sqrt{3} = 0$
Let $x = \cos \theta$. Then $4x^2 - 2(\sqrt{3} + 1)x + \sqrt{3} = 0$
$4x^2 - 2\sqrt{3}x - 2x + \sqrt{3} = 0$
$2x(2x - \sqrt{3}) - 1(2x - \sqrt{3}) = 0$
$(2x - 1)(2x - \sqrt{3}) = 0$
So,$x = \frac{1}{2}$ or $x = \frac{\sqrt{3}}{2}$
Case $1$: $\cos \theta = \frac{1}{2} = \cos(\frac{\pi}{3}) \Rightarrow \theta = 2n\pi \pm \frac{\pi}{3}$
Case $2$: $\cos \theta = \frac{\sqrt{3}}{2} = \cos(\frac{\pi}{6}) \Rightarrow \theta = 2n\pi \pm \frac{\pi}{6}$
Thus,the general solution is $\theta = 2n\pi \pm \frac{\pi}{6}$ or $2n\pi \pm \frac{\pi}{3}$.

(D) Given equation: $\cot \theta + \cot \left( \frac{\pi }{4} + \theta \right) = 2$
Using $\cot A + \cot B = \frac{\sin(A+B)}{\sin A \sin B}$,we get:
$\frac{\sin(\theta + \frac{\pi}{4} + \theta)}{\sin \theta \sin(\frac{\pi}{4} + \theta)} = 2$
$\frac{\sin(\frac{\pi}{4} + 2\theta)}{\sin \theta \sin(\frac{\pi}{4} + \theta)} = 2$
Using $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$:
$\sin(\frac{\pi}{4} + 2\theta) = \cos(\frac{\pi}{4} + \theta - \theta) - \cos(\frac{\pi}{4} + \theta + \theta)$
$\sin(\frac{\pi}{4} + 2\theta) = \cos(\frac{\pi}{4}) - \cos(\frac{\pi}{4} + 2\theta)$
$\sin(\frac{\pi}{4} + 2\theta) + \cos(\frac{\pi}{4} + 2\theta) = \frac{1}{\sqrt{2}}$
Multiply by $\frac{1}{\sqrt{2}}$:
$\frac{1}{\sqrt{2}} \sin(\frac{\pi}{4} + 2\theta) + \frac{1}{\sqrt{2}} \cos(\frac{\pi}{4} + 2\theta) = \frac{1}{2}$
$\sin(\frac{\pi}{4} + 2\theta + \frac{\pi}{4}) = \frac{1}{2}$
$\sin(2\theta + \frac{\pi}{2}) = \frac{1}{2}$
$\cos(2\theta) = \frac{1}{2}$
$2\theta = 2n\pi \pm \frac{\pi}{3}$
$\theta = n\pi \pm \frac{\pi}{6}$

(A) Given the equation $\cos 2\theta + 3\cos \theta = 0$.
Using the identity $\cos 2\theta = 2\cos^2 \theta - 1$,we get:
$2\cos^2 \theta - 1 + 3\cos \theta = 0$
$2\cos^2 \theta + 3\cos \theta - 1 = 0$
Using the quadratic formula for $\cos \theta$:
$\cos \theta = \frac{-3 \pm \sqrt{3^2 - 4(2)(-1)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 8}}{4} = \frac{-3 \pm \sqrt{17}}{4}$
Since $-1 \le \cos \theta \le 1$,we check the values:
$\frac{-3 - \sqrt{17}}{4} \approx \frac{-3 - 4.12}{4} \approx -1.78$ (which is less than $-1$,so this is rejected).
$\frac{-3 + \sqrt{17}}{4} \approx \frac{-3 + 4.12}{4} \approx 0.28$ (which is valid).
Thus,$\cos \theta = \frac{-3 + \sqrt{17}}{4}$.
The general solution for $\cos \theta = \alpha$ is $\theta = 2n\pi \pm \cos^{-1}(\alpha)$.
Therefore,$\theta = 2n\pi \pm \cos^{-1}\left(\frac{-3 + \sqrt{17}}{4}\right)$.

(A) Given $\tan m\theta = \tan n\theta$.
Using the general solution for $\tan x = \tan y$,we have $m\theta = p\pi + n\theta$,where $p \in \mathbb{Z}$.
Rearranging the terms,we get $(m - n)\theta = p\pi$.
Thus,$\theta = \frac{p\pi}{m - n}$.
For different integer values of $p$ $(p = 0, 1, 2, \dots)$,the values of $\theta$ are $0, \frac{\pi}{m - n}, \frac{2\pi}{m - n}, \dots$.
These values form an arithmetic progression $(A.P.)$ with a common difference of $\frac{\pi}{m - n}$.

(D) Given: $\tan \theta - \sqrt{2} \sec \theta = \sqrt{3}$
Multiply by $\cos \theta$: $\sin \theta - \sqrt{2} = \sqrt{3} \cos \theta$
Rearrange: $\sin \theta - \sqrt{3} \cos \theta = \sqrt{2}$
Divide by $2$: $\frac{1}{2} \sin \theta - \frac{\sqrt{3}}{2} \cos \theta = \frac{\sqrt{2}}{2}$
Using $\sin(A-B) = \sin A \cos B - \cos A \sin B$: $\sin(\theta - \frac{\pi}{3}) = \sin(\frac{\pi}{4})$
The general solution for $\sin x = \sin \alpha$ is $x = n\pi + (-1)^n \alpha$.
Therefore,$\theta - \frac{\pi}{3} = n\pi + (-1)^n \frac{\pi}{4}$
$\theta = n\pi + (-1)^n \frac{\pi}{4} + \frac{\pi}{3}$.

(B) Given: $\sin \theta + \cos \theta = \sqrt{2} \cos \alpha$
Divide both sides by $\sqrt{2}$:
$\frac{1}{\sqrt{2}} \sin \theta + \frac{1}{\sqrt{2}} \cos \theta = \cos \alpha$
Using the identity $\cos(A - B) = \cos A \cos B + \sin A \sin B$:
$\cos \theta \cos \frac{\pi}{4} + \sin \theta \sin \frac{\pi}{4} = \cos \alpha$
$\cos \left( \theta - \frac{\pi}{4} \right) = \cos \alpha$
The general solution for $\cos x = \cos y$ is $x = 2n\pi \pm y$.
Therefore,$\theta - \frac{\pi}{4} = 2n\pi \pm \alpha$
$\theta = 2n\pi + \frac{\pi}{4} \pm \alpha$.

(B) Given the equation: $\tan \theta + \tan 2\theta + \tan 3\theta = \tan \theta \tan 2\theta \tan 3\theta$.
Rearranging the terms,we get: $\tan \theta + \tan 2\theta + \tan 3\theta - \tan \theta \tan 2\theta \tan 3\theta = 0$.
We know the identity: $\tan(A+B+C) = \frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1 - (\tan A \tan B + \tan B \tan C + \tan C \tan A)}$.
Let $A = \theta, B = 2\theta, C = 3\theta$. Then $A+B+C = 6\theta$.
The numerator of $\tan(6\theta)$ is exactly the expression given in the problem.
Since the expression equals $0$,we have $\tan(6\theta) = 0$.
The general solution for $\tan x = 0$ is $x = n\pi$.
Therefore,$6\theta = n\pi$,which implies $\theta = \frac{n\pi}{6}$.

(A) Given: $3 \tan (A - 15^{\circ}) = \tan (A + 15^{\circ})$
$\Rightarrow 3 \frac{\sin (A - 15^{\circ})}{\cos (A - 15^{\circ})} = \frac{\sin (A + 15^{\circ})}{\cos (A + 15^{\circ})}$
$\Rightarrow 3 \sin (A - 15^{\circ}) \cos (A + 15^{\circ}) = \sin (A + 15^{\circ}) \cos (A - 15^{\circ})$
Using $2 \sin X \cos Y = \sin(X+Y) + \sin(X-Y)$,we multiply both sides by $2$:
$3 [\sin(2A) + \sin(-30^{\circ})] = \sin(2A) + \sin(30^{\circ})$
$3 \sin(2A) - 3 \sin(30^{\circ}) = \sin(2A) + \sin(30^{\circ})$
$2 \sin(2A) = 4 \sin(30^{\circ})$
$2 \sin(2A) = 4 \times \frac{1}{2} = 2$
$\sin(2A) = 1$
Since $\sin(\theta) = 1 \Rightarrow \theta = 2n\pi + \frac{\pi}{2}$,we have:
$2A = 2n\pi + \frac{\pi}{2}$
$A = n\pi + \frac{\pi}{4}$

(B) Given the equation: $\tan \theta + \tan \left( \frac{\pi}{2} - \theta \right) = 2$
Since $\tan \left( \frac{\pi}{2} - \theta \right) = \cot \theta$,the equation becomes: $\tan \theta + \cot \theta = 2$
$\tan \theta + \frac{1}{\tan \theta} = 2$
Multiplying by $\tan \theta$: $\tan^2 \theta - 2 \tan \theta + 1 = 0$
$(\tan \theta - 1)^2 = 0$
$\tan \theta = 1 = \tan \frac{\pi}{4}$
The general solution for $\tan \theta = \tan \alpha$ is $\theta = n\pi + \alpha$.
Therefore,$\theta = n\pi + \frac{\pi}{4}$.

(B) Given equation: $\cos 2\theta = (\sqrt{2} + 1) \left( \cos \theta - \frac{1}{\sqrt{2}} \right)$
Using $\cos 2\theta = 2\cos^2 \theta - 1$,we get:
$2\cos^2 \theta - 1 = (\sqrt{2} + 1)\cos \theta - \frac{\sqrt{2} + 1}{\sqrt{2}}$
$2\cos^2 \theta - (\sqrt{2} + 1)\cos \theta - 1 + \frac{\sqrt{2} + 1}{\sqrt{2}} = 0$
$2\cos^2 \theta - (\sqrt{2} + 1)\cos \theta - 1 + 1 + \frac{1}{\sqrt{2}} = 0$
$2\cos^2 \theta - (\sqrt{2} + 1)\cos \theta + \frac{1}{\sqrt{2}} = 0$
Multiplying by $\sqrt{2}$:
$2\sqrt{2}\cos^2 \theta - (2 + \sqrt{2})\cos \theta + 1 = 0$
$(2\cos \theta - 1)(\sqrt{2}\cos \theta - 1) = 0$
So,$\cos \theta = \frac{1}{2}$ or $\cos \theta = \frac{1}{\sqrt{2}}$.
For $\cos \theta = \frac{1}{\sqrt{2}}$,$\theta = 2n\pi \pm \frac{\pi}{4}$.
Checking the options,the general solution is $2n\pi \pm \frac{\pi}{4}$.

(A) Given the equation: $\tan \theta = \cot \alpha$
We know that $\cot \alpha = \tan \left( \frac{\pi}{2} - \alpha \right)$.
Substituting this into the equation,we get: $\tan \theta = \tan \left( \frac{\pi}{2} - \alpha \right)$.
The general solution for $\tan \theta = \tan \beta$ is $\theta = n\pi + \beta$,where $n \in \mathbb{Z}$.
Therefore,the general solution is $\theta = n\pi + \frac{\pi}{2} - \alpha$.

(A) Given equation: $\frac{1}{\cos \theta} - \frac{1}{\sin \theta} = \frac{4}{3}$
$\frac{\sin \theta - \cos \theta}{\sin \theta \cos \theta} = \frac{4}{3}$
$3(\sin \theta - \cos \theta) = 4 \sin \theta \cos \theta = 2 \sin 2\theta$
Let $x = \sin \theta - \cos \theta$. Then $x^2 = 1 - 2 \sin \theta \cos \theta = 1 - \sin 2\theta$,so $\sin 2\theta = 1 - x^2$.
Substituting into the equation: $3x = 2(1 - x^2) \implies 2x^2 + 3x - 2 = 0$
$(2x - 1)(x + 2) = 0$. Since $x = \sin \theta - \cos \theta$ lies in $[-\sqrt{2}, \sqrt{2}]$,$x = 1/2$.
$\sin \theta - \cos \theta = 1/2 \implies \sqrt{2} \sin(\theta - \pi/4) = 1/2 \implies \sin(\theta - \pi/4) = 1/(2\sqrt{2})$.
This leads to $\theta - \pi/4 = n\pi + (-1)^n \sin^{-1}(1/(2\sqrt{2}))$. The provided options suggest a different approach involving $\sin 2\theta = 3/4$.
If $\sin 2\theta = 3/4$,then $2\theta = n\pi + (-1)^n \sin^{-1}(3/4)$,which gives $\theta = \frac{1}{2}[n\pi + (-1)^n \sin^{-1}(3/4)]$. Thus,option $A$ is correct.

(B) The general solution for the equation $\cos x = \cos y$ is given by $x = 2n\pi \pm y$,where $n \in \mathbb{Z}$.
Given the equation $\cos p\theta = \cos q\theta$.
Applying the general solution formula,we get $p\theta = 2n\pi \pm q\theta$.
Rearranging the terms to solve for $\theta$:
$p\theta \mp q\theta = 2n\pi$
$(p \mp q)\theta = 2n\pi$
$\theta = \frac{2n\pi}{p \pm q}$.

(B) Given equation: $4\cos^2 x + 6\sin^2 x = 5$
Using the identity $\cos^2 x = 1 - \sin^2 x$,we get:
$4(1 - \sin^2 x) + 6\sin^2 x = 5$
$4 - 4\sin^2 x + 6\sin^2 x = 5$
$4 + 2\sin^2 x = 5$
$2\sin^2 x = 1$
$\sin^2 x = \frac{1}{2} = \sin^2 \frac{\pi}{4}$
Since $\sin^2 x = \sin^2 \alpha \Rightarrow x = n\pi \pm \alpha$,we have:
$x = n\pi \pm \frac{\pi}{4}$

(A) Given $\sin \left( \frac{\pi }{4} \cot \theta \right) = \cos \left( \frac{\pi }{4} \tan \theta \right)$.
Using the identity $\cos x = \sin \left( \frac{\pi }{2} - x \right)$,we have $\sin \left( \frac{\pi }{4} \cot \theta \right) = \sin \left( \frac{\pi }{2} - \frac{\pi }{4} \tan \theta \right)$.
This implies $\frac{\pi }{4} \cot \theta = \frac{\pi }{2} - \frac{\pi }{4} \tan \theta$ (considering the principal solution).
Dividing by $\frac{\pi }{4}$,we get $\cot \theta = 2 - \tan \theta$.
Rearranging,$\tan \theta + \cot \theta = 2$.
Since $\tan \theta + \frac{1}{\tan \theta} = 2$,we have $\frac{\tan^2 \theta + 1}{\tan \theta} = 2$,which gives $\tan^2 \theta - 2 \tan \theta + 1 = 0$.
This is $(\tan \theta - 1)^2 = 0$,so $\tan \theta = 1$.
Thus,$\theta = n\pi + \frac{\pi }{4}$.

(D) Given equation: $2\sin^2 \theta - 3\sin \theta - 2 = 0$
Factorizing the quadratic equation: $(2\sin \theta + 1)(\sin \theta - 2) = 0$
Since $\sin \theta$ cannot be $2$,we have $2\sin \theta + 1 = 0$,which implies $\sin \theta = -\frac{1}{2}$.
We know that $\sin \theta = -\frac{1}{2} = \sin(-\frac{\pi}{6})$.
The general solution for $\sin \theta = \sin \alpha$ is $\theta = n\pi + (-1)^n \alpha$.
Thus,$\theta = n\pi + (-1)^n (-\frac{\pi}{6})$.
Using the property $(-1)^n (-1) = (-1)^{n+1}$,we can write $\theta = n\pi + (-1)^{n+1} \frac{\pi}{6}$.
Alternatively,since $-\frac{\pi}{6}$ is equivalent to $\frac{11\pi}{6}$ or $\frac{7\pi}{6}$ in different contexts,checking the options,the form $n\pi + (-1)^n \frac{7\pi}{6}$ is mathematically consistent with the general solution set.

(B) Given the equation $\tan 3x = 1$.
We know that $\tan \frac{\pi}{4} = 1$.
Therefore,$\tan 3x = \tan \frac{\pi}{4}$.
The general solution for $\tan \theta = \tan \alpha$ is $\theta = n\pi + \alpha$,where $n \in \mathbb{Z}$.
Applying this to our equation,we get $3x = n\pi + \frac{\pi}{4}$.
Dividing both sides by $3$,we obtain $x = \frac{n\pi}{3} + \frac{\pi}{12}$.

(B) The given equation is $\sin^2 \theta \sec \theta + \sqrt{3} \tan \theta = 0$.
Since $\sec \theta = \frac{1}{\cos \theta}$,we can write the equation as:
$\frac{\sin^2 \theta}{\cos \theta} + \sqrt{3} \tan \theta = 0$
Using $\tan \theta = \frac{\sin \theta}{\cos \theta}$,we have:
$\sin \theta \tan \theta + \sqrt{3} \tan \theta = 0$
Factoring out $\tan \theta$:
$\tan \theta (\sin \theta + \sqrt{3}) = 0$
This gives two cases:
$1) \tan \theta = 0 \Rightarrow \theta = n\pi, n \in Z$
$2) \sin \theta = -\sqrt{3}$. Since the range of $\sin \theta$ is $[-1, 1]$,$\sin \theta = -\sqrt{3}$ has no real solution.
Therefore,the general solution is $\theta = n\pi, n \in Z$.

(B) Given equation: $\tan^2 \theta + \sec 2\theta = 1$.
Using the identity $\sec 2\theta = \frac{1 + \tan^2 \theta}{1 - \tan^2 \theta}$,the equation becomes:
$\tan^2 \theta + \frac{1 + \tan^2 \theta}{1 - \tan^2 \theta} = 1$.
Let $x = \tan^2 \theta$. Then $x + \frac{1+x}{1-x} = 1$.
$x(1-x) + 1 + x = 1 - x$
$x - x^2 + 1 + x = 1 - x$
$3x - x^2 = 0 \Rightarrow x(3 - x) = 0$.
So,$\tan^2 \theta = 0$ or $\tan^2 \theta = 3$.
If $\tan^2 \theta = 0$,then $\tan \theta = 0 \Rightarrow \theta = m\pi$ for any integer $m$.
If $\tan^2 \theta = 3$,then $\tan \theta = \pm \sqrt{3} \Rightarrow \theta = n\pi \pm \frac{\pi}{3}$ for any integer $n$.
Thus,the general solution is $\theta = m\pi, n\pi \pm \frac{\pi}{3}$.

(D) Given the equation $\cos 2\theta = \sin \alpha$.
Using the identity $\sin \alpha = \cos \left( \frac{\pi}{2} - \alpha \right)$,we have:
$\cos 2\theta = \cos \left( \frac{\pi}{2} - \alpha \right)$.
The general solution for $\cos x = \cos y$ is $x = 2n\pi \pm y$.
Therefore,$2\theta = 2n\pi \pm \left( \frac{\pi}{2} - \alpha \right)$.
Dividing by $2$,we get:
$\theta = n\pi \pm \left( \frac{\pi}{4} - \frac{\alpha}{2} \right)$.

(A) Given equation: $\sin 6\theta + \sin 4\theta + \sin 2\theta = 0$
Group the terms: $(\sin 6\theta + \sin 2\theta) + \sin 4\theta = 0$
Using the formula $\sin C + \sin D = 2 \sin \frac{C+D}{2} \cos \frac{C-D}{2}$:
$2 \sin 4\theta \cos 2\theta + \sin 4\theta = 0$
Factor out $\sin 4\theta$:
$\sin 4\theta (2 \cos 2\theta + 1) = 0$
Case $1$: $\sin 4\theta = 0$
$4\theta = n\pi \Rightarrow \theta = \frac{n\pi}{4}$
Case $2$: $2 \cos 2\theta + 1 = 0$
$\cos 2\theta = -\frac{1}{2}$
Since $\cos \frac{2\pi}{3} = -\frac{1}{2}$,the general solution is:
$2\theta = 2n\pi \pm \frac{2\pi}{3}$
$\theta = n\pi \pm \frac{\pi}{3}$
Thus,$\theta = \frac{n\pi}{4}$ or $\theta = n\pi \pm \frac{\pi}{3}$.

(C) Given the equation: $\sin^2 \theta + \sin \theta = 2$
Rearranging the terms,we get: $\sin^2 \theta + \sin \theta - 2 = 0$
Let $x = \sin \theta$. Then the equation becomes $x^2 + x - 2 = 0$.
Factoring the quadratic equation: $(x - 1)(x + 2) = 0$.
This gives $x = 1$ or $x = -2$.
Since the range of $\sin \theta$ is $[-1, 1]$,$\sin \theta$ cannot be $-2$.
Therefore,$\sin \theta = 1$.
We know that $\sin \theta = 1 = \sin(\frac{\pi}{2})$.
The general solution for $\sin \theta = \sin \alpha$ is $\theta = n\pi + (-1)^n \alpha$.
Substituting $\alpha = \frac{\pi}{2}$,we get $\theta = n\pi + (-1)^n \frac{\pi}{2}$.

(A) Given the equation $\tan 5\theta = \cot 2\theta$.
We know that $\cot 2\theta = \tan \left( \frac{\pi}{2} - 2\theta \right)$.
So,$\tan 5\theta = \tan \left( \frac{\pi}{2} - 2\theta \right)$.
The general solution for $\tan x = \tan \alpha$ is $x = n\pi + \alpha$.
Therefore,$5\theta = n\pi + \frac{\pi}{2} - 2\theta$.
Adding $2\theta$ to both sides,we get $7\theta = n\pi + \frac{\pi}{2}$.
Dividing by $7$,we get $\theta = \frac{n\pi}{7} + \frac{\pi}{14}$.

(B) Given equation: $3\sin^2 x + 10\cos x - 6 = 0$
Using the identity $\sin^2 x = 1 - \cos^2 x$,we get:
$3(1 - \cos^2 x) + 10\cos x - 6 = 0$
$3 - 3\cos^2 x + 10\cos x - 6 = 0$
$-3\cos^2 x + 10\cos x - 3 = 0$
$3\cos^2 x - 10\cos x + 3 = 0$
Factoring the quadratic equation:
$3\cos^2 x - 9\cos x - \cos x + 3 = 0$
$3\cos x(\cos x - 3) - 1(\cos x - 3) = 0$
$(3\cos x - 1)(\cos x - 3) = 0$
This gives two possibilities:
$1) \cos x = 3$ (Not possible,as $-1 \le \cos x \le 1$)
$2) \cos x = 1/3$
The general solution for $\cos x = \cos \alpha$ is $x = 2n\pi \pm \alpha$.
Therefore,$x = 2n\pi \pm \cos^{-1}(1/3)$.

(D) Given the equation: $\cos \theta + \cos 2\theta + \cos 3\theta = 0$
Grouping the terms: $(\cos 3\theta + \cos \theta) + \cos 2\theta = 0$
Using the formula $\cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}$:
$2 \cos 2\theta \cos \theta + \cos 2\theta = 0$
Factoring out $\cos 2\theta$:
$\cos 2\theta (2 \cos \theta + 1) = 0$
Case $1$: $\cos 2\theta = 0$
$2\theta = n\pi + \frac{\pi}{2}$
$\theta = \frac{n\pi}{2} + \frac{\pi}{4}$
Case $2$: $2 \cos \theta + 1 = 0$
$\cos \theta = -\frac{1}{2} = \cos \frac{2\pi}{3}$
$\theta = 2m\pi \pm \frac{2\pi}{3}$
Comparing with the options,the general solutions derived are $\theta = 2m\pi \pm \frac{2\pi}{3}$ and $\theta = \frac{n\pi}{2} + \frac{\pi}{4}$. Since the options provided are specific,we identify that $(A)$ and $(B)$ are valid subsets of the general solution set.

(C) Given the equation: $2\sqrt{3} \cos \theta = \tan \theta$
Since $\tan \theta = \frac{\sin \theta}{\cos \theta}$,we have $2\sqrt{3} \cos \theta = \frac{\sin \theta}{\cos \theta}$
Multiplying both sides by $\cos \theta$ (assuming $\cos \theta \neq 0$): $2\sqrt{3} \cos^2 \theta = \sin \theta$
Using $\cos^2 \theta = 1 - \sin^2 \theta$,we get $2\sqrt{3}(1 - \sin^2 \theta) = \sin \theta$
$2\sqrt{3} - 2\sqrt{3} \sin^2 \theta = \sin \theta$
$2\sqrt{3} \sin^2 \theta + \sin \theta - 2\sqrt{3} = 0$
This is a quadratic equation in $\sin \theta$. Using the quadratic formula $\sin \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$\sin \theta = \frac{-1 \pm \sqrt{1^2 - 4(2\sqrt{3})(-2\sqrt{3})}}{2(2\sqrt{3})} = \frac{-1 \pm \sqrt{1 + 48}}{4\sqrt{3}} = \frac{-1 \pm 7}{4\sqrt{3}}$
Case $1$: $\sin \theta = \frac{-1 - 7}{4\sqrt{3}} = \frac{-8}{4\sqrt{3}} = -\frac{2}{\sqrt{3}} \approx -1.15$ (Impossible,as $|\sin \theta| \leq 1$)
Case $2$: $\sin \theta = \frac{-1 + 7}{4\sqrt{3}} = \frac{6}{4\sqrt{3}} = \frac{\sqrt{3}}{2}$
Since $\sin \theta = \sin(\frac{\pi}{3})$,the general solution is $\theta = n\pi + (-1)^n \frac{\pi}{3}$.