Trigonometric Equations

$\begin{aligned} & \text { } \cos 2 \theta=\cos \theta+\sin \theta \\ & \Rightarrow \quad\left(\cos ^2 \theta-\sin ^2 \theta\right)-(\cos \theta+\sin \theta)=0 \\ & \Rightarrow \quad(\cos \theta+\sin \theta)(\cos \theta-\sin \theta-1)=0 \\ & \Rightarrow \quad \cos \theta+\sin \theta=0 \\ & \text { or } \cos \theta-\sin \theta=1 \\ & \text { When } \cos \theta+\sin \theta=0 \text {, then } \tan \theta=-1 \\ & \text { then } \theta=n \pi+\frac{3 \pi}{4} \Rightarrow \theta=\frac{3 \pi}{4}, \frac{7 \pi}{4} \in[0,2 \pi] \\ & \text { and when } \cos \theta-\sin \theta=1 \text {, then } \sin \theta+(1-\cos \theta)=0 \\ & \text { and } \quad 2 \sin \frac{\theta}{2}\left(\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right)=0 \\ & \end{aligned}$

$\begin{aligned} &\text { and } \quad \sin \frac{\theta}{2}=0, \cos \frac{\theta}{2}+\sin \frac{\theta}{2}=0 \\ & \Rightarrow \quad \frac{\theta}{2}=n \pi, \tan \frac{\theta}{2}=-1 \\ & \Rightarrow \quad \theta=0,2 \pi \in[0,2 \pi] \text {, } \\ & \theta=\frac{6 \pi}{4} \in[0,2 \pi] \\ \end{aligned}$

Sum of all values of $\theta$

$=\frac{3 \pi}{4}+\frac{7 \pi}{4}+0+2 \pi+\frac{6 \pi}{4}=\frac{24 \pi}{4}$

$3 \operatorname{cosec} x=4 \sin x$

or $\sin ^2 x=3 / 4$

or $ \sin ^2 x=\sin ^2 \frac{\pi}{3} \Rightarrow x=n \pi \pm \frac{\pi}{3}, n \in Z$

or $x= \pm \frac{\pi}{3}$

${\cos ^3}x\,.\,\sin 2x = {{\cos 3x + 3\cos x} \over 4} \times \sin 2x$

$ = {1 \over 8}(\sin 5x - \sin x) + {3 \over 8}(\sin 3x + \sin x)$

$ = {1 \over 4}\sin x + {3 \over 8}\sin 3x + {1 \over 8}\sin 5x$

Here, $n = 5,{a_1} = {1 \over 4},{a_2} = 0,{a_3} = {1 \over 8},{a_4} = 0,{a_5} = {1 \over 8}$

$\left| {\cot x} \right| = \cot x + {1 \over {\sin x}}$

Let $\cot x > 0 \Rightarrow \cot x = \cot x + {1 \over {\sin x}} = 0$

$ \Rightarrow {1 \over {\sin x}} = 0$ which is not possible.

Let $\cot x \le 0 \Rightarrow - \cot x = \cot x + {1 \over {\sin x}}$

$ \Rightarrow - 2\cot x = {1 \over {\sin x}} \Rightarrow \cos x = - {1 \over 2} \Rightarrow x = {{2\pi } \over 3},{{8\pi } \over 3}$

$\therefore$ The number of solutions is 2.

$ \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} $

For real roots, discriminant,

$D = {b^2} - 4ac \ge 0$

$ = {\cos ^2}\beta - 4(\cos \beta - 1)sin\beta \ge 0$

$ = {\cos ^2}\beta + 4(1 - \cos \beta )\sin \beta \ge 0$

So, sin$\beta$ should be > 0. [$\because$ ${\cos ^2}\beta \ge 0,\,1 - \cos \beta \ge 0$]

$ \Rightarrow \beta \in (0,\pi )$.

We have,

${5 \over 4}{\cos ^2}2x + {\cos ^4}x + {\sin ^4}x + {\cos ^6}x + {\sin ^6}x = 2$

$ \Rightarrow {5 \over 4}{\cos ^2}2x + {({\cos ^2}x + {\sin ^2}x)^2} - 2{\sin ^2}x{\cos ^2}x + {({\cos ^2}x + {\sin ^2}x)^3} - 3{\sin ^2}x{\cos ^2}x({\sin ^2}x + {\cos ^2}x) = 2$

$ \Rightarrow {5 \over 4}{\cos ^2}2x + 1 - 2{\sin ^2}x{\cos ^2}x + 1 - 3{\sin ^2}x{\cos ^2}x = 2$

$ \Rightarrow {5 \over 4}{\cos ^2}2x - {1 \over 2}{\sin ^2}2x - {3 \over 4}{\sin ^2}2x = 0$

$ \Rightarrow {5 \over 4}({\cos ^2}2x - {\sin ^2}2x) = 0$

$ \Rightarrow \cos 4x = 0$

$ \Rightarrow 4x = 2n\pi \, \pm {\pi \over 2},n \in Z$

$ \Rightarrow x = (4n \pm 1){\pi \over 8},n \in Z$

$ \Rightarrow x = {\pi \over 8},{{3\pi } \over 8},{{5\pi } \over 8},{{7\pi } \over 8},{{9\pi } \over 8},{{11\pi } \over 8},{{13\pi } \over 8},{{15\pi } \over 8}$

So, number of solution = 8