Trigonometric Equations

Given the equation:

$ \sin 9\theta = \sin \theta $

This can be rewritten as:

$ \sin 9\theta - \sin \theta = 0 $

Using the identity $ \sin C - \sin D = 2 \cos \left(\frac{C+D}{2}\right) \sin \left(\frac{C-D}{2}\right) $, we have:

$ 2 \cos \left(\frac{9\theta + \theta}{2}\right) \sin \left(\frac{9\theta - \theta}{2}\right) = 0 $

Simplifying, this becomes:

$ 2 \cos 5\theta \sin 4\theta = 0 $

This implies either:

$ \cos 5\theta = 0 $

$ \sin 4\theta = 0 $

For $ \cos 5\theta = 0 $, we have:

$ 5\theta = (2n + 1) \frac{\pi}{2} $

This gives:

$ \theta = (2n + 1) \frac{\pi}{10} $

For $ \sin 4\theta = 0 $, we have:

$ 4\theta = n\pi $

This gives:

$ \theta = \frac{n\pi}{4} $

Calculating these solutions within the interval $[0, 2\pi]$:

From $ \theta = (2n+1) \frac{\pi}{10} $:

$ \theta = \frac{\pi}{10}, \frac{3\pi}{10}, \frac{\pi}{2}, \frac{7\pi}{10}, \frac{9\pi}{10}, \frac{11\pi}{10}, \frac{13\pi}{10}, \frac{3\pi}{2}, \frac{17\pi}{10}, \frac{19\pi}{10} $

From $ \theta = \frac{n\pi}{4} $:

$ \theta = 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi, \frac{5\pi}{4}, \frac{3\pi}{2}, \frac{7\pi}{4}, 2\pi $

The total number of solutions counting common terms $ \frac{\pi}{2} $ and $ \frac{3\pi}{2} $ only once, is:

$ 17 $

Given, $2 \cos x\left(4 \sin \left(\frac{\pi}{4}+x\right) \sin \left(\frac{\pi}{4}-x\right)-1\right)=1$

$\begin{array}{cc} \Rightarrow & 2 \cos x\left(2 \cos (2 x)-2 \cos \left(\frac{\pi}{2}\right)-1\right)=1 \\ \Rightarrow & 2 \cos x\left(4 \cos ^2 x-3\right)=1 \\ \Rightarrow & 4 \cos ^3 x-3 \cos x=\frac{1}{2} \\ \Rightarrow & \cos 3 x=\frac{1}{2} \\ \Rightarrow & 3 x=\frac{\pi}{3}, \frac{5 \pi}{3}, \frac{7 \pi}{3} \\ & x=\frac{\pi}{9}, \frac{5 \pi}{9}, \frac{7 \pi}{9} \end{array}$

$\therefore$ Number of solutions $(n)=3$ and sum of solutions

$\begin{aligned} \quad(S) & =\frac{\pi}{9}+\frac{5 \pi}{9}+\frac{7 \pi}{9}=\frac{13 \pi}{9} \\ \therefore(n, S) & =\left(3, \frac{13 \pi}{9}\right) \end{aligned}$

$2\cos \theta [\cos 120^\circ + \cos 2\theta ] = 1$

$ \Rightarrow 2\cos \theta \left( { - {1 \over 2} + 2{{\cos }^2}\theta - 1} \right) = 1$

$ \Rightarrow 4{\cos ^3}\theta - 3\cos - 1 = 0$

$ \Rightarrow \cos 3\theta = 1 = \cos 0$

$ \Rightarrow 3\theta = 2n\pi $ or $\theta = {{2n\pi } \over 3},n \in z$

Given the values so that 2n does not exceed 18.

Hence, the sum $ = {{2\pi } \over 3}\sum\limits_1^9 {n = {{2\pi } \over 3} \times {{9(9 + 1)} \over 2} = 30\pi } $

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

$ = {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.

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