Trigonometric Equations

$ \begin{aligned} & \text { } \alpha^2+12+3 \sin (a+b \alpha)+6 \alpha=0 \\ & 3 \sin (a+b \alpha)=-\alpha^2-12-6 \alpha \end{aligned} $

Since, $\alpha \in$ real

$ \begin{array}{ll} & -3 \leq 3 \sin (a+b \alpha) \leq 3 \\ & -3 \leq-\alpha^2-12-6 \alpha \leq 3 \\ \Rightarrow \quad & -\alpha^2-12-6 \alpha \geq-3 \\ \Rightarrow \quad & -\alpha^2-6 \alpha-9 \geq 0 \\ \Rightarrow \quad & (\alpha+3)^2 \leq 0 \\ \Rightarrow \quad & \alpha=-3 \end{array} $

$ \begin{aligned} & \text { Also, } \alpha^2+6 \alpha+15 \geq 0 \\ & \Rightarrow \quad \alpha \in R \end{aligned} $

So, $\quad \alpha=-3$

$ \begin{aligned} & \Rightarrow \quad 3 \sin (a+b \alpha)+9+12-18=0 \\ & \Rightarrow \quad \sin (a+b \alpha)=-1 \\ & \Rightarrow \quad a+b \alpha=\frac{3 \pi}{2} \end{aligned} $

So, $\cos (a+b \alpha)=0$

$ \begin{aligned} & \cot A+\cot B+\tan A+\tan B=\cot A \cot B -\tan A \tan B \\ & \Rightarrow \quad \tan B+\tan A+\tan ^2 A \tan B +\tan ^2 B \tan A \\ &= 1-\tan ^2 A \tan ^2 B \end{aligned} $

$ \begin{aligned} & \Rightarrow \tan B(1+\tan A \tan B) +\tan A(1+\tan A \tan B) \\ & \quad=(1+\tan A \tan B)(1-\tan A \tan B) \\ & \Rightarrow \frac{\tan A+\tan B}{1-\tan A \tan B}=1 \\ & \Rightarrow \quad \tan (A+B)=1 \\ & \Rightarrow \quad A+B=\frac{\pi}{4} \end{aligned} $

$ \therefore \quad \sin (A+B)=\frac{1}{\sqrt{2}} $

$ \begin{aligned} & \text { } \tan ^2 x+3 \cot ^2 x=2 \sec ^2 x \\ & \qquad \begin{array}{l} \tan ^2 x+\frac{3}{\tan ^2 x}=2\left(1+\tan ^2 x\right) \\ \left(x \neq 0, \pi, 2 \pi, \frac{\pi}{2}, \frac{3 \pi}{2}\right) \end{array} \end{aligned} $

$ \begin{array}{ll} \text { Let } & t=\tan ^2 x \\ & t+\frac{3}{t}=2(1+t) \\ \Rightarrow & t^2+3=2 t+2 t^2 \\ \Rightarrow & t^2+2 t-3=0 \\ \Rightarrow & (t+3)(t-1)=0 \\ \Rightarrow & t=1 \\ \Rightarrow & \tan ^2 x=1 \\ \Rightarrow & x=n \pi \pm \frac{\pi}{4} \\ \Rightarrow & x=\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4} \end{array} $

$ \text { } \begin{aligned} & 2 \sin \theta+3 \cos \theta=2 \\ & \Rightarrow(2 \sin \theta+3 \cos \theta)^2=4 \\ & \Rightarrow 4 \sin ^2 \theta+9 \cos ^2 \theta+12 \sin \theta \cos \theta=4 \\ & \Rightarrow 5 \cos ^2 \theta+12 \sin \theta \cos \theta=0 \\ & \Rightarrow \cos \theta(5 \cos \theta+12 \sin \theta)=0 \\ & \Rightarrow \tan \theta=\frac{-5}{12} \\ & \sin \theta=\frac{-5}{13}, \cos \theta=\frac{12}{13} \\ & \therefore \sin \theta+\cos \theta=\frac{-5}{13}+\frac{12}{13}=\frac{7}{13} \end{aligned} $

$ \begin{aligned} &\text { } 2 \sin x \sin 3 x \sin 5 x+\cos 4 x \sin 5 x=0\\ &\begin{aligned} & =\sin 5 x(2 \sin x \sin 3 x+\cos 4 x)=0 \\ & =\sin 5 x(2 \sin x \sin 3 x+\cos x \cos 3 x-\sin x \sin 3 x)=0 \\ & =\sin 5 x(\cos x \cos 3 x+\sin x \cdot \sin 3 x)=0 \\ & =\sin 5 x \cdot \cos 2 x=0 \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} \therefore & =\sin 5 x=0 & & \cos 2 x=0 \\ & =5 x=n \pi & & 2 x=(2 k+1) \frac{\pi}{2} \\ x & =\frac{n \pi}{5} & & x=(2 k+1) \frac{\pi}{4} \\ & n \in I & & k \in I \\ \because & x \in(-\pi, \pi) & & \\ & -\pi < \frac{n \pi}{5}<\pi & & -\pi < (2 k+1) \frac{\pi}{4} < \pi \\ & -5 < n < 5 & & -4 < 2 k+1 < 4 \\ & & \Rightarrow & \frac{-5}{2} < k < \frac{3}{2} \end{aligned}\\ &\text { Number of values of } x \text { is }\\ &9+4=13 \end{aligned} $

We have

$ \begin{array}{r} \cos \alpha+\cos \beta+\cos \gamma=0 \\ =\sin \alpha+\sin \beta+\sin \gamma \end{array} $

Let $a=\cos \alpha+i \sin \alpha$

$ \begin{gathered} b=\cos \beta+i \sin \beta \\ c=\cos \gamma+i \sin \gamma \\ \therefore a+b+c=(\cos \alpha+\cos \beta+\cos \gamma) \\ +i(\sin \alpha+\sin \beta+\sin \gamma) \end{gathered} $

$ a+b+c=0 $

and $a^{-1}+b^{-1}+c^{-1}=0$

$ \begin{aligned} & \therefore a b+b c+c a=0 \\ & \Rightarrow \cos (\alpha+\beta)+\cos (\beta+\gamma) \\ & \quad+\cos (\gamma+\alpha)=0 \end{aligned} $

Now $a^2+b^2+c^2=(a+b+c)^2$

$ -2(a b+b c+c a) $

$ \therefore \sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma=0 $

$\sin ^2 \theta+2 \cos ^2 \theta-\sqrt{3} \sin \theta \cos \theta=2$

$ \begin{aligned} & \Rightarrow 1+\cos ^2 \theta-\sqrt{3} \sin \theta \cos \theta=2 \\ & \Rightarrow 1+\cos 2 \theta-\sqrt{3} \sin 2 \theta-2=0 \\ & \Rightarrow \cos 2 \theta-\sqrt{3} \sin 2 \theta=1 \end{aligned} $

Divide by $\sqrt{1^2+(-\sqrt{3})^2}=\sqrt{4}=2$ on both sides

$ \Rightarrow \frac{1}{2} \cos 2 \theta-\frac{\sqrt{3}}{2} \sin 2 \theta=\frac{1}{2} $

$ \begin{aligned} & \Rightarrow \quad \cos \left(2 \theta+\frac{\pi}{3}\right)=\frac{1}{2} \\ & \Rightarrow \quad 2 \theta+\frac{\pi}{3}=2 n \pi \pm \frac{\pi}{3}, n \in Z \end{aligned} $

If $2 \theta+\frac{\pi}{3}=2 n \pi+\frac{\pi}{3} \Rightarrow \theta=n \pi, n \in Z$

If $2 \theta+\frac{\pi}{3}=2 n \pi-\frac{\pi}{3} \Rightarrow \theta=n \pi-\frac{\pi}{3}, n \in Z$

For $\theta=n \pi$, if $n=0, \theta=0 \in(-\pi, \pi)$

For $\theta=n \pi-\frac{\pi}{3}$, if $n=0, \theta=\frac{-\pi}{3} \in(-\pi, \pi)$

if $n=1, \theta=\pi-\frac{\pi}{3}=\frac{2 \pi}{3} \in(-\pi, \pi)$

∴ Solutions are $0, \frac{-\pi}{3}, \frac{2 \pi}{3}$

⇒ Number of solutions $=3$

$ \begin{aligned} &\text { Given equation is }\\ &\begin{array}{ll} & 1+\cos x+\cos ^2 x+\ldots+\infty=4+2 \sqrt{3} \\ \Rightarrow & \frac{1}{1-\cos x}=4+2 \sqrt{3} \\ \Rightarrow & 1-\cos x=\frac{1}{4+2 \sqrt{3}} \times \frac{4-2 \sqrt{3}}{4-2 \sqrt{3}} \\ \Rightarrow & 1-\cos x=\frac{4-2 \sqrt{3}}{4} \Rightarrow 1-\cos x=1-\frac{\sqrt{3}}{2} \\ \Rightarrow & \cos x=\frac{\sqrt{3}}{2}=\cos \frac{\pi}{6} \\ \Rightarrow & x=2 n \pi \pm \frac{\pi}{6} \Rightarrow x=\frac{12 n \pi \pm \pi}{6} \\ \Rightarrow & x=(12 n \pm 1) \frac{\pi}{6} \end{array} \end{aligned} $

Given, $\sin ^2 x+b \sin x+c=0$ and

$ \alpha+\beta=\frac{\pi}{2} $

Let $y=\sin x$, thus we get

$ y^2+b y+c=0 $

The roots of this equation are $\sin \alpha$ and $\sin \beta$.

So, sum of the roots $=\sin \alpha+\sin \beta=-\frac{b}{1}$

Product of the roots $=\sin \alpha \cdot \sin \beta=c$

Since, $\alpha+\beta=\frac{\pi}{2}$

$ \Rightarrow \quad \beta=\frac{\pi}{2}-\alpha $

So, $\sin \alpha+\sin \left(\frac{\pi}{2}-\alpha\right)=-b$

$ \begin{aligned} \Rightarrow \sin \alpha+\cos \alpha=-b & \\ & {\left[\because \sin \left(\frac{\pi}{2}-\theta\right)=\cos \theta\right] } \end{aligned} $

$\Rightarrow(\sin \alpha+\cos \alpha)^2=(-b)^2$

$ \Rightarrow \sin ^2 \alpha+\cos ^2 \alpha+2 \sin \alpha \cos \alpha=(-b)^2 $

$\Rightarrow 1+2 c=b^2$\left[\because \sin ^2 \theta+\cos ^2 \theta=1\right]$

$ \Rightarrow b^2-1=2 c $

Given equation

$ \begin{aligned} & \sqrt{6-5 \cos x+7 \sin ^2 x}-\cos x=0 \\ \Rightarrow & \sqrt{6-5 \cos x+7-7 \cos ^2 x}-\cos x=0 \\ \Rightarrow & \sqrt{13-5 \cos x-7 \cos ^2 x}-\cos x=0 \end{aligned} $

Squaring both sides, we get

$ \begin{aligned} & \Rightarrow 13-5 \cos x-7 \cos ^2 x=\cos ^2 x \\ & \Rightarrow 13-5 \cos x-8 \cos ^2 x=0 \end{aligned} $

$ \Rightarrow 8 \cos ^2 x+5 \cos x+13=0 $

Let $y=\cos x$

$ \begin{aligned} & \Rightarrow 8 y^2+5 y-13=0 \\ & \qquad y=\frac{-5 \pm \sqrt{25-4(8)(-13)}}{2 \times 8} \\ & \quad=\frac{-5 \pm \sqrt{25+416}}{16}=\frac{-5 \pm 21}{16} \\ & \Rightarrow \quad y=\frac{-5+21}{16} \text { or } y=\frac{-5-21}{16} \\ & \Rightarrow \quad y=1 \text { or } y=-\frac{26}{16}=-\frac{13}{8} (not\,\, possible)\end{aligned} $

$ \therefore \quad \cos x=1 \Rightarrow \cos x=1 $

And $\sin ^2 x=1-\cos ^2 x=0 \Rightarrow \sin x=0$

So, $\tan x=\frac{\sin x}{\cos x}=\frac{0}{1}=0$

$ \sec x=\frac{1}{\cos x}=1 $

Now, $\tan x+\cot x=0+\frac{1}{0}=\infty$

$ \begin{aligned} & \cot x+\operatorname{cosec} x=\frac{1}{0}+\frac{1}{0}=\infty \\ & \tan x+\sec x=0+1=1 \\ & \sec x+\operatorname{cosec} x=1+\frac{1}{0}=\infty \end{aligned} $

Given the equations:

$ \sin \theta_{1} + \sin \theta_{2} = -\frac{21}{65} $

$ \cos \theta_{1} + \cos \theta_{2} = -\frac{27}{65} $

By squaring both equations and adding them, we have:

$ \cos^2 \theta_{1} + \cos^2 \theta_{2} + \sin^2 \theta_{1} + \sin^2 \theta_{2} + 2\left[\cos(\theta_{1}-\theta_{2})\right] $

This equates to:

$ = \left(\frac{21}{65}\right)^2 + \left(\frac{27}{65}\right)^2 $

Calculating further, we get:

$ 2 + 2\cos(\theta_{1}-\theta_{2}) = \frac{441 + 729}{65^2} $

Simplifying, we have:

$ 2\left[2\cos^2\left(\frac{\theta_{1}-\theta_{2}}{2}\right)\right] = \frac{1170}{65^2} $

Therefore:

$ \cos\left(\frac{\theta_{1}-\theta_{2}}{2}\right) = \frac{-3}{\sqrt{130}} $

Given that $\frac{\theta_1-\theta_2}{2}$ lies in the second quadrant, the result confirms $\cos$ will be negative in this quadrant. Hence:

$ \cos\left(\frac{\theta_{1}-\theta_{2}}{2}\right) = -\frac{3}{\sqrt{130}} $

To solve this problem, let's consider the solution sets of the given equations step by step.

Set A:

We start with the equation: $\cos^2 x = \cos^2 \frac{\pi}{6}$.

Since $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$, we have $\cos^2 \frac{\pi}{6} = \frac{3}{4}$.

The solutions to this equation are:

$ x = n\pi \pm \frac{\pi}{6}, \quad \text{where } n \in \mathbb{Z} $

Set B:

We consider the equation: $\cos^2 x = \log_{16} P$.

Given $ P + \frac{16}{P} = 10 $, finding $P$ involves solving the quadratic equation $p^2 - 10p + 16 = 0$.

Factoring gives: $(p - 8)(p - 2) = 0$, so $p = 8$ or $p = 2$.

Thus, $\cos^2 x = \log_{16} 8$ or $\cos^2 x = \log_{16} 2$.

For $p = 8$, $\log_{16} 8 = \frac{3}{4}$ (since $\frac{3}{4} = \cos^2\left(\frac{\pi}{6}\right)$).

For $p = 2$, $\log_{16} 2 = \frac{1}{4}$.

Hence, $\cos^2 x = \frac{3}{4}$ aligns with $\cos^2 x = \cos^2 \frac{\pi}{6}$.

For $\cos^2 x = \frac{1}{4}$, the solutions are:

$ x = 2n\pi \pm \frac{\pi}{3} \quad \text{and} \quad x = 2n\pi \pm \frac{2\pi}{3}, \quad \text{where } n \in \mathbb{Z} $

Finding $B - A$:

Set $B$ includes solutions for both $\cos^2 x = \frac{1}{4}$ and $\cos^2 x = \frac{3}{4}$.

Set $A$ only includes solutions for $\cos^2 x = \frac{3}{4}$.

Therefore, the difference $B-A$ consists of solutions only when $\cos^2 x = \frac{1}{4}$, which are:

$ B - A = \left\{x \in \mathbb{R} \,|\, x = 2n\pi \pm \frac{\pi}{3}, \, 2n\pi \pm \frac{2\pi}{3}, \, n \in \mathbb{Z} \right\} $

Given the problem:

$ \tan A + \tan B + \cot A + \cot B = \tan A \tan B - \cot A \cot B $

We can express the trigonometric functions in terms of sine and cosine. That gives:

$ \frac{\sin A}{\cos A} + \frac{\sin B}{\cos B} + \frac{\cos A}{\sin A} + \frac{\cos B}{\sin B} = \frac{\sin A}{\cos A} \cdot \frac{\sin B}{\cos B} - \frac{\cos A}{\sin A} \cdot \frac{\cos B}{\sin B} $

Simplifying each term, we get:

$ \frac{\sin^2 A + \cos^2 A}{\sin A \cos A} + \frac{\sin^2 B + \cos^2 B}{\sin B \cos B} = \frac{\sin^2 A \sin^2 B - \cos^2 A \cos^2 B}{\cos A \cos B \sin A \sin B} $

Since $ \sin^2 \theta + \cos^2 \theta = 1 $, this reduces to:

$ \frac{1}{\sin A \cos A} + \frac{1}{\sin B \cos B} = \frac{\sin^2 A \sin^2 B - \left( 1 - \sin^2 A \right) \left( 1 - \sin^2 B \right)}{\cos A \cos B \sin A \sin B} $

$ \Rightarrow \frac{1}{2}(\sin 2A + \sin 2B) = \sin^2 A + \sin^2 B - 1 $

Using the sine addition and subtraction formulas, $\sin(2x) = 2\sin x \cos x$, transform the equation:

$ \sin(A+B) \cos(A-B) = \sin^2 A - \cos^2 B $

Simplifying further:

$ \sin(A+B) \cos(A-B) = -\cos(A+B) \cos(A-B) $

This implies:

$ \tan(A+B) = -1 \Rightarrow A + B = 135^\circ $

Given the constraint $0^\circ < A+B < 270^\circ$, the valid solution is:

$ A+B = 135^\circ $

Given, $ 4-3 \cos ^{2} \theta=5 \sin \theta \cos \theta $

$ 4 \sec ^{2} \theta-3=5 \tan \theta \quad\left[\because\right. $ divide by $ \left.\cos ^{2} \theta\right] $ $ 4+4 \tan ^{2} \theta-3=5 \tan \theta $

$ \left[\because 1+\tan ^{2} \theta=\sec ^{2} \theta\right] $

$ 4 \tan ^{2} \theta-5 \tan \theta+1=0 $

$ 4 \tan ^{2} \theta-4 \tan \theta-\tan \theta+1=0 $

$ (4 \tan \theta-1)(\tan \theta-1) $

$ \tan \theta=\frac{1}{4}, 1 $

If $ \tan \theta=1 $, then $ \theta=45^{\circ} $

Now, put the value of $ \theta $ in option,

$ \therefore 1+\sin ^{2} \theta=1+\frac{1}{2}=\frac{3}{2}=3 \cos ^{2} \theta $

Hence, $ 1+\sin ^{2} \theta=3 \cos ^{2} \theta $

We have, $\cos ^2 2 x+\sin ^2 3 x=1$

$ \begin{aligned} & \Rightarrow \quad 1-\sin ^2 2 x+\sin ^3 3 x=1 \\ & \Rightarrow \quad \sin ^2 3 x=\sin ^2 2 x \Rightarrow 3 x=n x \pm 2 x \end{aligned} $

On taking + ve sign, we get

$ x=n \pi, n \in Z $

On taking - ve sign, we get

$ \begin{array}{ll} & 3 x=n \pi-2 x, n \in Z \\ & 5 x=n \pi, n \in Z \\ & x=\frac{n \pi}{5}, n \in Z \\ & x \in\left\{\left.\frac{n \pi}{5} \right\rvert\, n \in Z\right\} \\ \therefore \quad & x \in\left\{\left.\frac{n \pi}{5} \right\rvert\, n \in Z\right\} \cup\{n \pi \mid n \in Z\} \\ & =\left\{\left.\frac{n \pi}{5} \right\rvert\, n \in Z\right\} \end{array} $

Hence, solution set is $\left\{x / x=\frac{n \pi}{5}, n \in Z\right\}$

We have,

$ \begin{aligned} & f(x)=2 \cos (3 x+4)-3 \tan (2 x-3) + 5 \sin (5 x)-7 \end{aligned} $

For $\cos (3 x+4)$, period is $\frac{2 \pi}{3}$

For $\tan (2 x-3)$, period is $\frac{\pi}{2}$

For $\sin (5 x)$, period is $\frac{2 \pi}{5}$

To find overall period of $f(x)$, we need the LCM of $\frac{2 \pi}{3}, \frac{\pi}{2}, \frac{2 \pi}{5}$. or LCM of $\frac{40 \pi}{60}, \frac{30 \pi}{60}, \frac{24 \pi}{60}$

So, the overall period $k$ of the function

$ \begin{aligned} & f(x) \text { is } \frac{120 \pi}{60}=2 \pi \\\\ & \Rightarrow \tan \frac{k}{3}=\tan \frac{2 \pi}{3}=-\sqrt{3} \end{aligned} $

We have,

$ \begin{aligned} & \sin 7 \theta-\sin 3 \theta=\sin 4 \theta \\\\ & 2 \cos \left(\frac{7 \theta+3 \theta}{2}\right) \sin \left(\frac{7 \theta-3 \theta}{2}\right)=\sin 4 \theta \\\\ & 2 \cos 5 \theta \sin 2 \theta=2 \sin 2 \theta \cos 2 \theta \\\\ & \sin 2 \theta(\cos 5 \theta-\cos 2 \theta)=0 \\\\ & \sin 2 \theta\left[-\sin \left(\frac{7 \theta}{2}\right) \sin \left(\frac{3 \theta}{2}\right)\right]=0 \end{aligned} $

$ \begin{aligned} & \sin 2 \theta \cdot \sin \frac{7 \theta}{2} \cdot \sin \frac{3 \theta}{2}=0 \\\\ & \sin 2 \theta=0 \Rightarrow \theta=\frac{\pi}{2} \text { in }(0, \pi) \\\\ & \sin \frac{7 \theta}{2}=0 \Rightarrow \theta=\frac{2 \pi}{7}, \frac{4 \pi}{7}, \frac{6 \pi}{7} \text { is }(0, \pi) \\\\ & \sin \frac{3 \theta}{2}=0 \Rightarrow \theta=\frac{2 \pi}{3} \end{aligned} $

Hence, total number of solutions of equation $=5$

$ \begin{aligned} & \text { Given, } \\ & \begin{array}{l} (\sqrt{3}-1) \sin \theta+(\sqrt{3}+1) \cos \theta=2 \\ \text { Let } \quad \sqrt{3}-1=r \sin \alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} \\ & \text { and } \quad \sqrt{3}+1=r \cos \alpha \\ & \begin{aligned} \therefore & r^2\left(\sin ^2 \alpha+\cos ^2 \alpha\right) \\ & =(\sqrt{3}-1)^2+(\sqrt{3}+1)^2 \end{aligned} \\ & \Rightarrow \quad r^2=3+1-2 \sqrt{3}+3+1+2 \sqrt{3} \\ & \Rightarrow \quad r^2=8 \\ & \Rightarrow \quad r=2 \sqrt{2} \text { and } \\ & \tan \alpha=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{1-1 / \sqrt{3}}{1+1 / \sqrt{3}} \\ & \quad=\tan \left(\frac{\pi}{4}-\frac{\pi}{6}\right)=\tan \left(\frac{\pi}{12}\right) \\ & \Rightarrow \alpha=\frac{\pi}{12} \end{aligned} $

From Eq. (i), we have

$ \begin{array}{ll} \Rightarrow & r \cos (\theta-\alpha)=2 \\ \Rightarrow & 2 \sqrt{2} \cos (\theta-\alpha)=2 \\ \Rightarrow & \cos (\theta-\alpha)=\frac{1}{\sqrt{2}} \\ \Rightarrow & \theta-\alpha=2 n \pi \pm \frac{\pi}{4} \\ \Rightarrow & \theta=2 n \pi \pm \frac{\pi}{4}+\frac{\pi}{12} \end{array} $

$ \begin{aligned} &\text { We have, } \tan \theta+5 \cot \theta=\sec \theta\\ &\begin{aligned} & \Rightarrow \frac{\sin \theta}{\cos \theta}+\frac{5 \cos \theta}{\sin \theta}=\frac{1}{\cos \theta^{\prime}}, \sin \theta \neq 0, \\ & \cos \theta \neq 0 \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow \frac{\sin ^2 \theta+5 \cos ^2 \theta}{\cos \theta \sin \theta}=\frac{1}{\cos \theta}, \sin \theta \neq 0 \\ & \cos \theta \neq 0 \\ & \Rightarrow \sin ^2 \theta+5\left(1-\sin ^2 \theta\right)=\sin \theta, \sin \theta \neq 0 \\ & \cos \theta \neq 0 \\ & \Rightarrow 4 \sin ^2 \theta+\sin \theta-5=0, \sin \theta \neq 0 \\ & \cos \theta \neq 0 \\ & \Rightarrow(4 \sin \theta+5)(\sin \theta-1)=0, \sin \theta \neq 0 \\ & \cos \theta \neq 0 \\ & \sin \theta=\frac{-5}{4}, 1, \sin \theta \neq 0, \cos \theta \neq 0 \\ & \Rightarrow \sin \theta=1, \sin \theta \neq 0, \cos \theta \neq 0 \\ & \quad[\because-1 \leq \sin \theta \leq 1] \end{aligned} $

$ \begin{aligned} & \Rightarrow \theta \in \phi \\ & {[\text { if } \sin \theta=0 \Rightarrow \cos \theta=0]} \end{aligned} $