Question

The number of solution(s) of $\cos 2\theta =\sin 4\theta$ and $\tan \theta =\cot 5\theta$ for $\theta \in (-\frac{\pi }{2},\frac{\pi }{2})$ is

Answer 1

Given: $\cos 2\theta =\sin 4\theta$ and $\tan \theta =\cot 5\theta$
Now,
$\cos 2\theta =\sin 4\theta \Rightarrow \cos 2\theta -2\sin 2\theta \cos 2\theta =0\Rightarrow \cos 2\theta (1-2\sin 2\theta )=0\Rightarrow \cos 2\theta =0,\sin 2\theta =\frac{1}{2}$
As $\theta \in (-\frac{\pi }{2},\frac{\pi }{2})\Rightarrow 2\theta \in (-\pi ,\pi )$
So,
$2\theta =-\frac{\pi }{2},\frac{\pi }{2},\frac{\pi }{6},\frac{5\pi }{6}\Rightarrow \theta =-\frac{\pi }{4},\frac{\pi }{4},\frac{\pi }{12},\frac{5\pi }{12}\cdots \left(1\right)$

Also,
$\tan \theta =\cot 5\theta \Rightarrow \frac{\sin \theta }{\cos \theta }-\frac{\cos 5\theta }{\sin 5\theta }=0\Rightarrow \frac{\sin \theta \sin 5\theta -\cos \theta \cos 5\theta }{\cos \theta \sin 5\theta }=0\Rightarrow -\frac{\cos 6\theta }{\cos \theta \sin 5\theta }=0\Rightarrow \cos 6\theta =0,\cos \theta \ne 0,\sin 5\theta \ne 0$
As $\theta \in (-\frac{\pi }{2},\frac{\pi }{2})\Rightarrow 6\theta \in (-3\pi ,3\pi )$
So,
$6\theta =-\frac{5\pi }{2},-\frac{3\pi }{2},-\frac{\pi }{2},\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2}\Rightarrow \theta =-\frac{5\pi }{12},-\frac{3\pi }{12},-\frac{\pi }{12},\frac{\pi }{12},\frac{3\pi }{12},\frac{5\pi }{12}\Rightarrow x=-\frac{5\pi }{12},-\frac{\pi }{4},-\frac{\pi }{12},\frac{\pi }{12},\frac{\pi }{4},\frac{5\pi }{12}\cdots \left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$, we get
$x=-\frac{\pi }{4},\frac{\pi }{4},\frac{\pi }{12},\frac{5\pi }{12}$

Hence, the number of solutions of the given equation in $(-\frac{\pi }{2},\frac{\pi }{2})$ is $4$.