Question

Number of solutions of the equation $|\sin 4\theta |=1$ in $\theta \in [0,2\pi ]$ is

Answer 1

$|\sin 4\theta |=1$
$\Rightarrow \sin 4\theta =\pm 1$

When $\sin 4\theta =1$
$\Rightarrow 4\theta =(4n+1)\frac{\pi }{2},n\in Z$
$\Rightarrow \theta =(4n+1)\frac{\pi }{8},n\in Z$
Number of solutions in $[0,2\pi ]$ are $4.$
i.e. $\{\frac{\pi }{8},\frac{5\pi }{8},\frac{9\pi }{8},\frac{13\pi }{8}\}$

When $\sin 4\theta =-1$
$\Rightarrow 4\theta =(4n-1)\frac{\pi }{2},n\in Z$
$\Rightarrow \theta =(4n-1)\frac{\pi }{8},n\in Z$
Number of solutions in $[0,2\pi ]$ are $4.$
i.e. $\{\frac{3\pi }{8},\frac{7\pi }{8},\frac{11\pi }{8},\frac{15\pi }{8}\}$

Hence, total number of solutions $=8$

Alternate Solution:
We know that $|\sin \theta |$ is periodic with period $\pi .$
So, $|\sin 4\theta |$ will be periodic with period $\frac{\pi }{4}.$
$\therefore$ Graph of $y=|\sin 4\theta |$ in $\theta \in [0,\pi ]$ is


So, number of solution in $[0,\pi /4]$ is $1.$
$\therefore$ Number of solution in $[0,2\pi ]$ is $8.$