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} $

$\tan \left(\frac{\pi}{4}+\alpha\right)=\tan ^3\left(\frac{\pi}{4}+\beta\right)$

let $x=\frac{\pi}{4}+\alpha, y=\frac{\pi}{4}+\beta$

$\therefore \tan x=\tan ^3 y$

Now, $\alpha+\beta=x+y-\frac{\pi}{2}$

and $\alpha-\beta=x-y$

then, $\tan (\alpha+\beta)=\tan \left(x+y-\frac{\pi}{2}\right)$

$ =-\cot (x+y) $

and $\cot (\alpha-\beta)=\cot (x-y)$

Thus, $\tan (\alpha+\beta) \cot (\alpha-\beta)$

$ =-\cot (x+y) \cdot \cot (x-y) $

and $\cot (x+y) \cdot \cot (x-y)$

$ \begin{aligned} & =\frac{\left(1-\tan ^4 y\right)\left(1+\tan ^4 y\right)}{\tan ^2 y\left(1+\tan ^2 y\right)\left(1-\tan ^2 y\right)} \\ & =\frac{1+\tan ^4 y}{\tan ^2 y}=\cot ^2 y+\tan ^2 y \end{aligned} $

Thus, $-\cot (\alpha-\beta) \tan (\alpha+\beta)$

$ =-\left(\cot ^2 y+\tan ^2 y\right) $

and $\tan y=\tan \left(\frac{\pi}{4}+\beta\right)=\frac{1+\tan \beta}{1-\tan \beta}$

$ \Rightarrow \cot y=\frac{1-\tan \beta}{1+\tan \beta} $

Therefore,

$ \begin{gathered} \tan ^2 y+\cot ^2 y=\frac{(1-\tan \beta)^4+(1+\tan \beta)^4}{\left(1-\tan ^2 \beta\right)^2} \\ =\frac{(1-\tan \beta)^4}{(1-\tan \beta)^2(1+\tan \beta)^2} +\frac{(1+\tan \beta)^4}{(1-\tan \beta)^2(1+\tan \beta)^2} \end{gathered} $

$ =\left(\frac{1-\tan \beta}{1+\tan \beta}\right)^2+\left(\frac{1+\tan \beta}{1-\tan \beta}\right)^2 $

$ =\frac{\left(1+\tan ^2 \beta-2 \tan \beta\right)^2+\left(1+\tan ^2 \beta+2 \tan \beta\right)^2}{\left(1-\tan ^2 \beta\right)^2} $

$ \begin{aligned} & =\frac{\left. 2\left(1+\tan ^4 \beta+4 \tan ^2 \beta\right)+2\left(\tan ^2 \beta\right. +2 \tan ^3 \beta+2 \tan \beta+\tan ^2 \beta-2 \tan ^3 \beta-2 \tan \beta\right)}{\left(1-\tan ^2 \beta\right)^2} \\ & =2\left[\frac{1+\tan ^4 \beta+4 \tan ^2 \beta+2 \tan ^2 \beta}{\left(1-\tan ^2 \beta\right)^2}\right] \\ & =2\left[\frac{\left(\tan ^4 \beta+2 \tan ^2 \beta+1\right)+(2 \tan \beta)^2}{\left(1-\tan ^2 \beta\right)^2}\right] \\ & =2\left(\sec ^2 2 \beta+\tan ^2 2 \beta\right) \end{aligned} $

Let $A=\sqrt{\sin ^2 x-\sin x+\frac{1}{2}}$

$ \sin ^2 x-\sin x+1 / 2 \geq 0 $

but minimum value of $\sin ^2 x-\sin x+1 / 2=1 / 4$

$ \begin{aligned} & \Rightarrow \sqrt{\sin ^2 x-\sin x+\frac{1}{2}} \geq 1 / 2 \\ & \because x \in[0,3] \Rightarrow x=\pi / 6,5 \pi / 6 \\ & \therefore B=2^{\sec ^2 y} \leq 2 \end{aligned} $

because $\sec ^2 y=\frac{1}{\cos ^2 y}$

$ \Rightarrow \quad|\cos x| \leq 1 $

and $\sec ^2 y \geq 1$

So, $B=2$ and $A=1 / 2$ is only possible solution

for $B=2 \sec ^2 y=1$ in $y \in[0,3]$

$\sec ^2 y=1$ at $y=0$

therefore, possible pairs are

$(\pi / 6,0)$ and $(5 \pi / 6,0)$

$\therefore$ Number of solutions are 2 .

$ \begin{aligned} &\text { Statement I }\\ &\begin{aligned} & 2 \sin ^2 \theta-\cos 2 \theta=0 \\ \Rightarrow & 2 \sin ^2 \theta-\left(1-2 \sin ^2 \theta\right)=0 \\ \Rightarrow & 4 \sin ^2 \theta=1 \\ \Rightarrow & \sin ^2 \theta=\frac{1}{4} \\ \Rightarrow & 2 \cos ^2 \theta-3 \sin \theta=0 \\ \Rightarrow & 2-2 \sin ^2 \theta-3 \sin \theta=0 \\ \Rightarrow & 2 \sin ^2 \theta+3 \sin \theta-2=0 \\ \Rightarrow & 2 \sin ^2 \theta+4 \sin \theta-\sin \theta-2=0 \\ \Rightarrow & (2 \sin \theta-1)(\sin \theta+2)=0 \\ & \sin \theta=\frac{1}{2} \end{aligned} \end{aligned} $

∴ Common solution

$ \theta=\frac{\pi}{6}, \frac{5 \pi}{6} $

Number of solutions 2.

Statement II

$ \begin{array}{ll} \Rightarrow & 2 \cos ^2 \theta-3 \sin \theta=0 \\ \Rightarrow & \sin \theta=\frac{1}{2} \\ \because & \theta \in[0, \pi] \\ \therefore & \theta=\frac{\pi}{6}, \frac{5 \pi}{6} \end{array} $

Number of solutions 2.

$ \begin{aligned} &\text { Given, }\\ &\begin{aligned} & 4 \cos 2 \theta \cos 3 \theta=\sec \theta \\ & \Rightarrow \quad 4 \cos \theta \cos 2 \theta \cdot \cos 3 \theta=1 \\ & \Rightarrow \quad 2 \cos 2 \theta(\cos 4 \theta+\cos 2 \theta)=1 \\ & \Rightarrow \quad 2 \cos 2 \theta \cos 4 \theta+2 \cos ^2 2 \theta=1 \\ & \Rightarrow \quad 2 \cos 2 \theta \cos 4 \theta+2 \cos ^2 2 \theta-1=0 \\ & \Rightarrow \quad 2 \cos 2 \theta \cos 4 \theta+\cos 4 \theta=0 \\ & \Rightarrow \quad \cos 4 \theta(2 \cos 2 \theta+1)=0 \\ & \text { either } \quad \cos 4 \theta=0 \\ & \Rightarrow \quad 4 \theta=(2 n+1) \frac{\pi}{2} \Rightarrow \theta=(2 n+1) \frac{\pi}{8} \\ & \Rightarrow \quad \cos 2 \theta=-\frac{1}{2} \Rightarrow \cos 2 \theta=\cos \frac{2 \pi}{3} \\ & \Rightarrow \quad 2 \theta=2 n \pi \pm \frac{2 \pi}{3} \Rightarrow \theta=n \pi \pm \frac{\pi}{3} \\ & \because \theta \in(0,2 \pi) \\ & \therefore \theta=\frac{\pi}{8}, \frac{3 \pi}{8}, \frac{5 \pi}{8}, \frac{7 \pi}{8}, \frac{9 \pi}{8}, \frac{11 \pi}{8}, \frac{13 \pi}{8}, \frac{15 \pi}{8}\rightarrow 8 \text { solutions } \\ & \text { and } \theta=\frac{\pi}{3}, \pi \pm \frac{\pi}{3}, 2 \pi-\frac{\pi}{3} \rightarrow 4 \text { solutions } \\ & \therefore \text { Total number of solution }=12 \end{aligned} \end{aligned} $

Given, expression is

$ \tan \left(\frac{2 \pi}{7}\right) \cdot \tan \left(\frac{4 \pi}{7}\right)+\tan \left(\frac{4 \pi}{7}\right) \cdot \tan \left(\frac{\pi}{7}\right)+\tan \left(\frac{\pi}{7}\right) \cdot \tan \left(\frac{2 \pi}{7}\right) $

Let $x=\frac{\pi}{7}, y=\frac{2 \pi}{7}$ and $z=\frac{4 \pi}{7}$

So, the expression becomes

$ \tan (y) \cdot \tan (z)+\tan (z) \cdot \tan (x)+\tan (x) \cdot \tan (y) $

Now, $x+y+z=\frac{\pi}{7}+\frac{2 \pi}{7}+\frac{4 \pi}{7}=\frac{7 \pi}{7}=\pi$

Since, $x+y+z=\pi$, we can use the identify

$ \begin{aligned} & \tan (x) \cdot \tan (y)+\tan (y) \cdot \tan (z)+\tan (z) \cdot \tan (x)=-7 \\ & \therefore \tan \left(\frac{\pi}{7}\right) \tan \left(\frac{2 \pi}{7}\right)+\tan \left(\frac{4 \pi}{7}\right) \cdot \tan \left(\frac{2 \pi}{7}\right)+\tan \left(\frac{4 \pi}{7}\right) \tan \left(\frac{\pi}{7}\right)=-7 \end{aligned} $

Given, $\cos x \sqrt{16 \sin ^2 x}=1$

Since, $16 \sin ^2 x=4|\sin x|$

$ \begin{aligned} & \text { So, } \cos x(4|\sin x|)=1 \\ & \Rightarrow \quad 2 \sin 2 x= \pm 1 \end{aligned} $

Also, $\sin 2 x= \pm \frac{1}{2}$

$ \begin{gathered} x=\frac{\pi}{12}, \frac{5 \pi}{12}, \frac{7 \pi}{12}, \frac{11 \pi}{12} \\ \therefore \quad \text { Sum }=\frac{\pi}{12}+\frac{5 \pi}{12}+\frac{7 \pi}{12}+\frac{11 \pi}{12}=2 \pi \end{gathered} $

$ \begin{aligned} & \text { } \sqrt{3} \cos \theta+\sin \theta>0 \\ & \Rightarrow \quad \frac{\sqrt{3}}{2} \cos \theta+\frac{1}{2} \sin \theta>0 \\ & \Rightarrow \quad \cos \frac{\pi}{6} \cos \theta+\sin \frac{\pi}{6} \sin \theta>0 \\ & \Rightarrow \quad \cos \left(\theta-\frac{\pi}{6}\right)>0 \\ & \because \quad \cos x>0 \\ & \Rightarrow \quad x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \\ & \therefore \quad \frac{-\pi}{2}<\theta-\frac{\pi}{6}<\frac{\pi}{2} \\ & \Rightarrow \quad-\frac{\pi}{2}+\frac{\pi}{6}<\theta<\frac{\pi}{2}+\frac{\pi}{6} \\ & \Rightarrow \quad-\frac{\pi}{3}<\theta<\frac{2 \pi}{3} \end{aligned} $

Hence, $\frac{-\pi}{3}<\theta<\frac{2 \pi}{3}$

$\because \sin x=-3 / 5$ and $\cos x=-4 / 5$

$\sin x$ and $\cos x$ both are negative

So, $x$ must be lie in 3rd quadrant

Therefore,

$ \begin{aligned} x & =(2 n+1) \pi+\theta \\ \tan \theta & =\frac{(-3 / 5)}{(-4 / 5)}=3 / 4 \\ \Rightarrow \quad \theta & =\tan ^{-1}(3 / 4) \end{aligned} $

Hence, $x=(2 n+1) \pi+\tan ^{-1}(3 / 4), n \in z$

$ \begin{aligned} &\text { Given, trigonometric equation is }\\ &\begin{aligned} & \sec x \cdot \cos 5 x+1=0 \\ & \Rightarrow \frac{\cos 5 x+\cos x}{\cos x}=0 \\ & \Rightarrow \quad 2 \cdot \cos 3 x \cdot \cos 2 x=0 \end{aligned} \end{aligned} $

$ \begin{array}{r} \Rightarrow \quad \cos 3 x=0 \text { or } \cos 2 x=0 \\ \Rightarrow \quad 3 x=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \frac{7 \pi}{2}, \frac{9 \pi}{2}, \frac{11 \pi}{2} \\ \binom{\because 0 \leq x \leq 2 \pi}{\therefore 0 \leq 3 x \leq 6 \pi} \end{array} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \quad x=\frac{\pi}{6}, \frac{3 \pi}{6}, \frac{5 \pi}{6}, \frac{7 \pi}{6}, \frac{9 \pi}{6}, \frac{11 \pi}{6} \\ & \text { Also, } 2 x=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \frac{7 \pi}{2}\binom{\because 0 \leq x \leq 2 \pi}{\therefore 0 \leq 2 x \leq 4 \pi} \\ & \Rightarrow \quad x=\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4} \end{aligned}\\ &\therefore \text { Total number of solutions }=10 \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } 2 \sin x-\cos 2 x=1 \\ \Rightarrow & 2 \sin x-\left(1-2 \sin ^2 x\right)=1 \\ \Rightarrow & 2 \sin x-1+2 \sin ^2 x=1 \\ \Rightarrow & 2 \sin ^2 x+2 \sin x-2=0 \\ \Rightarrow & \sin ^2 x+\sin x-1=0 \\ \Rightarrow & \sin x=\frac{-1 \pm \sqrt{1+4}}{2 \times 1} \\ \Rightarrow & \sin x=\frac{-1 \pm \sqrt{5}}{2} \\ \Rightarrow & \sin x=\frac{\sqrt{5}-1}{2} \\ \Rightarrow & 3-2 \sin ^2 x=3-2 \frac{(\sqrt{5}-1)^2}{4} \\ & =\frac{12-2(5+1-2 \sqrt{5})}{4} \\ & =\frac{12-12+4 \sqrt{5}}{4}=\sqrt{5} \end{aligned} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \cos x+\cos 3 x=\sin x+\sin 3 x \\ & \Rightarrow 2 \cos 2 x \cdot \cos x=2 \sin 2 x \cdot \cos x \\ & \Rightarrow \cos x(\cos 2 x-\sin 2 x)=0 \\ & \text { but given } x \neq(2 n+1) \frac{\pi}{4} \\ & \Rightarrow \quad \cos 2 x-\sin 2 x=0 \\ & \Rightarrow \quad \sin 2 x=\cos 2 x \\ & \Rightarrow \quad \tan 2 x=1 \\ & \Rightarrow \quad \tan 2 x=\tan \frac{\pi}{4} \\ & \therefore \quad 2 x=n \pi+\frac{\pi}{4} \Rightarrow x=\frac{n \pi}{2}+\frac{\pi}{8} \end{aligned} \end{aligned} $

$\sin 2 x+\cos 4 x=2 x \in[-\pi, \pi]$

$ \begin{aligned} & \Rightarrow \quad \sin 2 x+1-2 \sin ^2 2 x=2 \\ & \Rightarrow \quad 2 \sin ^2 2 x-\sin 2 x+1=0 \\ & \Rightarrow \quad \sin 2 x=\frac{1 \pm \sqrt{1-8}}{2} \end{aligned} $

Not possible

Number of solution $=0$

We have,

$ \begin{aligned} & \begin{array}{r} \cos \theta+\cos 2 \theta-\sqrt{3}(\sin \theta+\sin 2 \theta)+1=0 \\ \Rightarrow \quad \cos \theta+2 \cos ^2 \theta-\sqrt{3} \sin \theta -2 \sqrt{3} \sin \theta \cos \theta=0 \\ \Rightarrow \quad(\cos \theta-\sqrt{3} \sin \theta)+\left(2 \cos ^2 \theta\right. -2 \sqrt{3} \sin \theta \cos \theta)=0 \end{array} \\ & \begin{array}{r} \Rightarrow 1(\cos \theta-\sqrt{3} \sin \theta) +2 \cos \theta(\cos \theta-\sqrt{3} \sin \theta)=0 \end{array} \\ & \Rightarrow \quad(\cos \theta-\sqrt{3} \sin \theta)(2 \cos \theta+1)=0 \\ & \Rightarrow \cos \theta-\sqrt{3} \sin \theta=0 \text { or } 2 \cos \theta=-1 \\ & \begin{aligned} \tan \theta=\frac{1}{\sqrt{3}} \text { or } \cos \theta=-\frac{1}{2}=\cos \frac{2 \pi}{3} \\ \quad \theta=n \pi+\frac{\pi}{6} \quad \text { or } \theta=2 n \pi+\frac{2 \pi}{3} \\ \theta=\frac{\pi}{6}, \frac{7 \pi}{6} \quad \text { or } \theta=\frac{2 \pi}{3}, \frac{4 \pi}{3} \end{aligned} \end{aligned} $

$\because$ Total number of solution in the interval $(0,2 \pi)$ is 4 .

We have,

$ 2 \sin ^2 \theta-3 \cos ^2 \theta=\sin \theta \cos \theta $

Dividing both sides by $\cos ^2 \theta$, we get $\cos \theta \neq 0$

$ \begin{aligned} & \Rightarrow \quad 2 \tan ^2 \theta-3=\tan \theta \\ & \Rightarrow \quad 2 \tan ^2 \theta-\tan \theta-3=0 \\ & \Rightarrow \quad(2 \tan \theta-3)(\tan \theta+1)=0 \\ & \because \quad \tan \theta=\frac{3}{2} \text { and } \tan \theta=-1 \end{aligned} $

Clearly, in the interval ( $-\pi, \pi$ ) $\tan \theta=\frac{3}{2}$ gives two solution and $\tan \theta=-1$ gives two solution $\therefore$ Total number of solution $=4$

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 $=5$

Given the equation:

$ \cot \frac{x}{2} - \cot x = \operatorname{cosec} \frac{x}{2} $

We can express this equation using trigonometric identities:

$ \Rightarrow \frac{\cos \frac{x}{2}}{\sin \frac{x}{2}} - \frac{1}{\sin x} = \operatorname{cosec} \frac{x}{2} $

$ \Rightarrow \left(\frac{\cos \frac{x}{2} - 1}{\sin \frac{x}{2}}\right) = \frac{\cos x}{\sin x} $

$ \Rightarrow \frac{\cos \frac{x}{2} - 1}{\sin \frac{x}{2}} = \frac{\cos x}{2 \sin \frac{x}{2} \cos \frac{x}{2}} $

$ \Rightarrow 2(\cos \frac{x}{2} - 1) = \frac{\cos x}{\cos \frac{x}{2}} $

$ \Rightarrow 2 \cos^2 \frac{x}{2} - 2 \cos \frac{x}{2} = \cos x $

Using the identity $\cos x = 2 \cos^2 \frac{x}{2} - 1$:

$ \Rightarrow 2 \cos^2 \frac{x}{2} - 2 \cos \frac{x}{2} = 2 \cos^2 \frac{x}{2} - 1 $

$ \Rightarrow \cos \frac{x}{2} = \frac{1}{2} = \cos \frac{\pi}{3} $

This implies:

$ \Rightarrow \frac{x}{2} = 2n\pi \pm \frac{\pi}{3} $

Thus, we find the general solution for $x$:

$ \Rightarrow x = 4n\pi \pm \frac{2\pi}{3} $

$ \begin{aligned} &\text { Given that }\\ &\begin{array}{ll} & 8^{1+\cos ^2 x+\cos ^4 x+\ldots}=4^3 \\ \Rightarrow \quad & \left.2^{3\left(1+\cos ^2 x+\cos ^4 x+\ldots\right.}\right)=2^6 \\ \Rightarrow & 3\left(1+\cos ^2 x+\cos ^4 x+\ldots .\right)=6 \\ \Rightarrow \quad & 1+\cos ^2 x+\cos ^4 x+\ldots=2 \\ \Rightarrow \quad & \cos ^2 x+\cos ^4 x+\ldots .=1 \\ \because & \text { Sum of GP of infinite terms }=\frac{a}{1-r} \\ \therefore \quad & \frac{\cos ^2 x}{1-\cos ^2 x}=1 \Rightarrow \cos ^2 x=1-\cos ^2 x \\ \Rightarrow \quad & 2 \cos ^2 x=1 \Rightarrow \cos ^2 x=\frac{1}{2} \\ \Rightarrow \quad & \quad x= \pm \frac{\pi}{4}, \pm \frac{3 \pi}{4} \end{array} \end{aligned} $

To solve the equation $\tan x + \tan 2x - \tan 3x = 0$, we start by understanding the identity given:

$ \tan x + \tan 2x = \tan 3x. $

Using trigonometric identities, we have:

$ \tan 2x = \frac{2 \tan x}{1 - \tan^2 x} $

$ \tan 3x = \frac{3 \tan x - \tan^3 x}{1 - 3 \tan^2 x}. $

Substituting these into the equation gives:

$ \tan x + \frac{2 \tan x}{1 - \tan^2 x} = \frac{3 \tan x - \tan^3 x}{1 - 3 \tan^2 x}. $

Simplifying this, we obtain:

$ \frac{\tan x (3 - \tan^2 x)}{1 - \tan^2 x} = \frac{\tan x (3 - \tan^2 x)}{1 - 3 \tan^2 x}. $

Clearing the denominators by multiplying through by $(1 - \tan^2 x)(1 - 3 \tan^2 x)$, we get:

$ \tan x(3 - \tan^2 x)(1 - 3 \tan^2 x) = \tan x(3 - \tan^2 x)(1 - \tan^2 x). $

Subtracting gives:

$ \tan x(3 - \tan^2 x)\{-2 \tan^2 x\} = 0. $

This implies:

$\tan^3 x = 0$ or

$3 - \tan^2 x = 0$.

Solving each equation separately:

$\tan^3 x = 0 \implies \tan x = 0 \implies x = n\pi, \, n \in \mathbb{Z}.$

$3 - \tan^2 x = 0 \implies \tan^2 x = 3 \implies \tan x = \pm \sqrt{3}.$

Solving $\tan x = \pm \sqrt{3}$, we get:

$\tan x = \sqrt{3} \implies x = n\pi + \frac{\pi}{3}, \, n \in \mathbb{Z}.$

$\tan x = -\sqrt{3} \implies x = n\pi - \frac{\pi}{3}, \, n \in \mathbb{Z}.$

Therefore, the general solution is:

$ \{x \mid x = n\pi \pm \frac{\pi}{3}, \, n\pi, \, n \in \mathbb{Z}\}. $

$ \begin{aligned} &\text { Given, } 2 \tan 2 \theta-\cot 2 \theta+1=0, \theta \in[0, x]\\ &\begin{aligned} & \Rightarrow \quad 2 \tan 2 \theta-\frac{1}{\tan 2 \theta}+1=0 \\ & \Rightarrow \quad 2 \tan ^2 2 \theta+\tan 2 \theta-1=0 \\ & \Rightarrow \quad 2 \tan ^2 2 \theta+2 \tan 2 \theta-\tan 2 \theta-1=0 \\ & \Rightarrow \quad 2 \tan 2 \theta(1+\tan 2 \theta)-(1+\tan 2 \theta)=0 \\ & \Rightarrow \quad(2 \tan 2 \theta-1)(1+\tan 2 \theta)=0 \\ & \Rightarrow \quad 2 \tan 2 \theta=1 \text { or } \tan 2 \theta=-1 \\ & \Rightarrow \quad \tan 2 \theta=\frac{1}{2} \\ & \text { or } \quad \tan 2 \theta=\tan \left(n \pi+\frac{3 \pi}{4}\right) \\ & \Rightarrow \quad 2 \theta=n \pi+\frac{3 \pi}{4} \Rightarrow \theta=\frac{n \pi}{2}+\frac{3 \pi}{8} \end{aligned} \end{aligned} $

Since, number of point of intersection is 4 .

Hence, number of solution is 4 .

Given the equation:

$ a \cos x + a \sin x + a = 2k + 1 $

we can rearrange it as:

$ a (\cos x + \sin x) = 2k + 1 - a $

Divide through by $a$ (since $a \neq 0$):

$ \cos x + \sin x = \frac{2k + 1 - a}{a} = \frac{2k + 1}{a} - 1 $

Recognizing that:

$ \cos x + \sin x = \sqrt{2}\left(\frac{1}{\sqrt{2}}\cos x + \frac{1}{\sqrt{2}}\sin x\right) $

Let $ \frac{1}{\sqrt{2}}\cos x + \frac{1}{\sqrt{2}}\sin x = \sin\left(x + \frac{\pi}{4}\right) $, then:

$ \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) = \frac{2k + 1}{a} - 1 $

So:

$ \sin\left(x + \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \left(\frac{2k + 1 - a}{a}\right) $

For $\sin(y)$ where $y$ is in the range $[-1, 1]$:

$ -1 \leq \frac{2k + 1 - a}{\sqrt{2} a} \leq 1 $

This implies:

$ -a\sqrt{2} \leq 2k + 1 - a \leq a\sqrt{2} $

Rearranging gives:

$ a - a\sqrt{2} \leq 2k + 1 \leq a\sqrt{2} + a $

Subtract 1 from each part:

$ a - a\sqrt{2} - 1 \leq 2k \leq a\sqrt{2} + a - 1 $

Finally, divide by 2:

$ \frac{a - a\sqrt{2} - 1}{2} \leq k \leq \frac{a\sqrt{2} + a - 1}{2} $

Thus, $k$ lies in the interval:

$ \left[\frac{a - \sqrt{2}a - 1}{2}, \frac{a + \sqrt{2}a - 1}{2}\right] $

Therefore, $k$ is within the interval $\left[\frac{a-1-\sqrt{2} a}{2}, \frac{a-1+\sqrt{2} a}{2}\right]$.

To solve the equation $\sin x + 3 \sin 3x + \sin 5x = 1$, we start by rewriting it as:

$ (\sin x + \sin 5x) + 3 \sin 3x = 0 $

Using the sum-to-product identities, the expression $\sin x + \sin 5x$ becomes:

$ 2 \sin 3x \cos 2x $

Thus, the equation becomes:

$ 2 \sin 3x \cos 2x + 3 \sin 3x = 0 $

Factoring out $\sin 3x$, we get:

$ \sin 3x (2 \cos 2x + 3) = 0 $

Which implies that:

$\sin 3x = 0$

$2 \cos 2x + 3 = 0$

For $\sin 3x = 0$, the solutions are:

$ 3x = n\pi \quad \Rightarrow \quad x = \frac{n\pi}{3}, \quad n \in \mathbb{Z} $

Next, for $2 \cos 2x + 3 = 0$, solving gives:

$ \cos 2x = -\frac{3}{2} $

However, since the cosine of an angle cannot exceed 1 or be less than -1, there are no solutions in real numbers for this equation.

Since the only valid solutions come from $\sin 3x = 0$, we consider the possible values of $\sin \alpha$ where $\alpha \in S$. Evaluating $\sin$ at these angles, we find:

$ \{\sin \alpha \, | \, \alpha \in S\} = \left\{\frac{\sqrt{3}}{2}, 0, \frac{-\sqrt{3}}{2}\right\} $

Given, $\sin ^2 \theta+3 \cos ^2 \theta=5 \sin \theta$

$ \begin{array}{ll} \Rightarrow & \sin ^2 \theta+3\left(1-\sin ^2 \theta\right)=5 \sin \theta \\ \Rightarrow & \sin ^2 \theta+3-3 \sin ^2 \theta=5 \sin \theta \\ \Rightarrow & -2 \sin ^2 \theta-5 \sin \theta+3=0 \\ \Rightarrow & 2 \sin ^2 \theta+5 \sin \theta-3=0 \\ \Rightarrow & \sin \theta=\frac{-6}{2}, \sin \theta=\frac{1}{2} \\ \therefore \quad & \sin \theta=\frac{1}{2} \quad[\because|\sin \theta| \leq 1] \\ \Rightarrow & \theta=n \pi+(-1)^n \frac{\pi}{6} \end{array} $

$\therefore$ The general solution is $n \pi+(-1)^n \frac{\pi}{6}$, $n \in Z$.

We are given two equations: $\sin x + \sin y = \sin(x+y)$ and $|x| + |y| = 1$.

Let's analyze the first equation:

$ \begin{aligned} &\sin x + \sin y = \sin(x+y) \\ \Rightarrow & 2 \sin \left(\frac{x+y}{2}\right) \cdot \cos \left(\frac{x-y}{2}\right) = 2 \sin \left(\frac{x+y}{2}\right) \cdot \cos \left(\frac{x+y}{2}\right) \\ \Rightarrow & \sin \left(\frac{x+y}{2}\right) \left[\cos \left(\frac{x-y}{2}\right) - \cos \left(\frac{x+y}{2}\right)\right] = 0 \\ \Rightarrow & \sin \left(\frac{x+y}{2}\right)\left[-2 \sin \frac{x}{2} \cdot \sin \left(-\frac{y}{2}\right)\right] = 0 \\ \Rightarrow & 2 \sin \left(\frac{x+y}{2}\right) \sin \frac{x}{2} \cdot \sin \frac{y}{2} = 0 \end{aligned} $

From this, we have three cases to consider:

Case 1: $\sin \left(\frac{x+y}{2}\right) = 0$

This implies:

$ \frac{x+y}{2} = 0 \Rightarrow x + y = 0 \Rightarrow x = -y $

Using the second equation $|x| + |y| = 1$:

$ |x| + |-x| = 1 \Rightarrow x = \pm \frac{1}{2} \quad \text{and} \quad y = \mp \frac{1}{2} $

Thus, the ordered pairs are $\left(\frac{1}{2}, -\frac{1}{2}\right)$ and $\left(-\frac{1}{2}, \frac{1}{2}\right)$.

Case 2: $\sin \frac{x}{2} = 0$

This gives:

$ \frac{x}{2} = 0 \Rightarrow x = 0 $

From $|x| + |y| = 1$:

$ 0 + |y| = 1 \Rightarrow |y| = 1 \Rightarrow y = \pm 1 $

So, the ordered pairs are $(0, 1)$ and $(0, -1)$.

Case 3: $\sin \frac{y}{2} = 0$

This implies:

$ \frac{y}{2} = 0 \Rightarrow y = 0 $

Using $|x| + |y| = 1$:

$ |x| + 0 = 1 \Rightarrow |x| = 1 \Rightarrow x = \pm 1 $

Therefore, the ordered pairs are $(1, 0)$ and $(-1, 0)$.

By combining all the cases, we have a total of 6 ordered pairs:

$\left(\frac{1}{2}, -\frac{1}{2}\right)$

$\left(-\frac{1}{2}, \frac{1}{2}\right)$

$(0, 1)$

$(0, -1)$

$(1, 0)$

$(-1, 0)$

$ \begin{aligned} &\text { We have, } 5 \sinh (x)-\cosh (x)=5\\ &\begin{aligned} & \Rightarrow \quad 5\left(\frac{e^x-e^{-x}}{2}\right)-\left(\frac{e^x+e^{-x}}{2}\right)=5 \\ & \Rightarrow \quad \frac{5 e^x-5 e^{-x}-e^x-e^{-x}}{2}=5 \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad \frac{4 e^x-6 e^{-x}}{2}=5 \Rightarrow 2 e^x-3 e^{-x}=5 \\ & \Rightarrow \quad 2 e^{2 x}-3=5 e^x \Rightarrow 2 e^{2 x}-5 e^x-3=0 \end{aligned} $

$ \text { Let, } e^x=y $

$ \begin{aligned} & \text { Then, } 2 y^2-5 y-3=0 \\ & \begin{aligned} y & =\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\ & =\frac{5 \pm \sqrt{(-5)^2-4 \times 2 \times(-3)}}{2 \times 2} \\ & =\frac{5 \pm \sqrt{25+24}}{4}=\frac{5 \pm \sqrt{49}}{4}=\frac{5 \pm 7}{4} \\ & =\frac{5+4}{4}, \frac{5-7}{4}=3,-\frac{1}{2} \end{aligned} \end{aligned} $

Thus, $e^x=3$ or $e^x=-\frac{1}{2}$

If $e^x=3$

Then,

$ \begin{aligned} \tanh x & =\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{3-\frac{1}{3}}{3+\frac{1}{3}} \\ & =\frac{8 / 3}{10 / 3}=\frac{8}{10}=\frac{4}{5} \end{aligned} $

If $e^x=-\frac{1}{2}$

$ \ \begin{aligned} { Then, } \tanh (x) & =\frac{e^x-e^{-x}}{e^x+e^{-x}} \\ & =\frac{-\frac{1}{2}+2}{-\frac{1}{2}-2}=\frac{\frac{3}{2}}{-\frac{5}{2}}=-\frac{3}{5} \end{aligned} $

We need to find the smallest positive value of $\theta$ for which the equation

$ \tan\left(\theta + 100^\circ\right) = \tan\left(\theta + 50^\circ\right) \tan(\theta) \tan\left(\theta - 50^\circ\right) $

holds true.

Step-by-Step Solution:

Start with the given equation:

$ \tan\left(\theta + 100^\circ\right) = \tan\left(\theta + 50^\circ\right) \tan(\theta) \tan\left(\theta - 50^\circ\right) $

Rearranging terms, we have:

$ \frac{\tan \left(\theta+100^\circ\right)}{\tan \left(\theta-50^\circ\right)} = \tan \left(\theta+50^\circ\right) \tan(\theta) $

Express tangent in terms of sine and cosine:

$ \frac{\sin \left(\theta+50^\circ\right)}{\cos \left(\theta+100^\circ\right) \cos \left(\theta-50^\circ\right)} = \frac{\sin \left(\theta+50^\circ\right) \sin \left(\theta-50^\circ\right)}{\cos \left(\theta+50^\circ\right) \cos(\theta)} $

Using the componendo and dividendo rule:

$ \frac{\sin \left(2 \theta+50^\circ\right)}{\sin \left(150^\circ\right)} = \frac{-\cos 50^\circ}{\cos \left(2 \theta+50^\circ\right)} $

Equating the terms, we proceed with the identities:

$ \sin \left(2 \theta+50^\circ\right) \cos \left(2 \theta+50^\circ\right) = -\cos 50^\circ \sin 150^\circ $

Multiply both sides by 2:

$ \sin \left(4 \theta+100^\circ\right) = -\sin 40^\circ $

Since $\sin(-x) = -\sin(x)$, we have:

$ \sin \left(4 \theta+100^\circ\right) = \sin(-40^\circ) $

The general solution for the sine equation is:

$ 4 \theta+100^\circ = n \cdot 180^\circ + (-1)^n \cdot (-40^\circ) $

For $n=1$, substitute:

$ 4 \theta + 100^\circ = 180^\circ + 40^\circ $

$ 4 \theta + 100^\circ = 220^\circ $

Solve for $\theta$:

$ 4 \theta = 220^\circ - 100^\circ $

$ 4 \theta = 120^\circ $

$ \theta = 30^\circ $

Thus, the smallest positive value of $\theta$ is $\boxed{30^\circ}$.

$4 \cos 2 x-4 \sqrt{3} \sin 2 x+\cos 3 x$

$ -\sqrt{3} \sin 3 x+\cos x-\sqrt{3} \sin x=0 $

Divide by 2 on both sides, we get

$ \begin{aligned} & \begin{array}{r} \begin{array}{r} {\left[\frac{1}{2} \cos 2 x-\frac{\sqrt{3}}{2} \sin 2 x\right]} \end{array} \\ +\left[\frac{1}{2} \cos 3 x-\frac{\sqrt{3}}{2} \sin 3 x\right] \\ + \end{array}+\left[\frac{1}{2} \cos x-\frac{\sqrt{3}}{2} \sin x\right]=0 \end{aligned} \begin{aligned} & \Rightarrow\left[\sin \frac{\pi}{6} \cos 2 x-\cos \frac{\pi}{6} \sin 2 x\right] \end{aligned} $

$\begin{aligned} & \quad+\left[\sin \frac{\pi}{6} \cos 3 x-\cos \frac{\pi}{6} \sin 3 x\right] \\ & \quad+\left[\sin \frac{\pi}{6} \cos x-\cos \frac{\pi}{6} \sin x\right]=0 \\ & \Rightarrow 4 \sin \left(\frac{\pi}{6}-2 x\right)+\sin \left(\frac{\pi}{6}-3 x\right) \\ & +\sin \left(\frac{\pi}{6}-x\right)=0 \\ & \Rightarrow 4 \sin \left(\frac{\pi}{6}-2 x\right)+2 \sin \left(\frac{\pi}{6}-2 x\right) \\ & \Rightarrow \\ & \Rightarrow 2 \sin \left(\frac{\pi}{6}-2 x\right)[\cos x+2]=0 \\ & \Rightarrow 2 \sin \left(\frac{\pi}{6}-2 x\right)=0 \\ & \\ & \quad \frac{\pi}{6}-2 x=-n \pi \\ & \\ & 2 x=n \pi+\frac{\cos x \neq-2]}{6} \Rightarrow x=\frac{n \pi}{2}+\frac{\pi}{12}\end{aligned}$

To solve the equation $2 \cos^2 x - 2 \tan x + 1 = 0$, we proceed with the following steps:

Start by rewriting the equation:

$ 2 \cos^2 x - 2 \tan x + 1 = 0 $

Using the identity $\cos^2 x = \frac{1 + \cos 2x}{2}$, substitute:

$ 2 \left(\frac{1 + \cos 2x}{2}\right) - 2 \tan x + 1 = 0 $

This simplifies to:

$ \cos 2x - 2 \tan x + 2 = 0 $

Express $\cos 2x$ in terms of $\tan x$:

$ \cos 2x = \frac{1 - \tan^2 x}{1 + \tan^2 x} $

Substitute this identity:

$ \frac{1 - \tan^2 x}{1 + \tan^2 x} - 2 \tan x + 2 = 0 $

Eliminate the fraction by multiplying through by $1 + \tan^2 x$:

$ 1 - \tan^2 x - 2 \tan x + 2 \tan^2 x + 2 = 0 $

Simplify the equation:

$ -2 \tan^3 x + \tan^2 x - 2 \tan x + 3 = 0 $

Rearrange and factor the polynomial, letting $ \tan x = y $:

$ 2y^3 - y^2 + 2y - 3 = 0 $

Factoring, we get:

$ (y - 1)(2y^2 + y + 3) = 0 $

Solve the factors:

$ y - 1 = 0 \quad \Rightarrow \quad y = 1 $

The quadratic $2y^2 + y + 3$ does not have real solutions (since its discriminant is negative).

Therefore, $\tan x = 1$, which implies:

$ x = n\pi + \frac{\pi}{4}, \quad n \in \mathbb{Z} $

The general solution to the equation is thus $x = n\pi + \frac{\pi}{4}$ where $n$ is any integer.