$
2(\cos x+\cos 2 x)+(1+2 \cos x) \sin 2 x=2 \sin x
$
Total no. of solutions in $[-\pi, \pi]$ is $\_

Question

$
2(\cos x+\cos 2 x)+(1+2 \cos x) \sin 2 x=2 \sin x
$
Total no. of solutions in $[-\pi, \pi]$ is $\_\_\_\_$

Professor PC Thomas Classes · Answer

Given : $2 \cos x+2 \cos 2 x+\sin 2 x+2 \cos x \sin 2 x=2 \sin x$
$$
\begin{aligned}
& 2 \cos x+2 \cos 2 x+2 \sin x \cos x+(\sin 3 x-\sin x)=0 \
& 2 \cos x+2 \cos 2 x+2 \sin x \cos x+2 \cos 2 x \sin x=0 \
& 2 \cos x(1+\sin x)+2 \cos 2 x(1+\sin x)=0 \
& (1+\sin x)(2 \cos x+2 \cos 2 x)=0 \
& 2(1+\sin x)(\cos x+\cos 2 x)=0 \
& 2(1+\sin x) 2 \cos \left(\frac{3 x}{2}ight) \cdot \cos \left(\frac{x}{2}ight)=0
\end{aligned}
$$
$$
\begin{gathered}
\Rightarrow \sin x=-1 \Rightarrow x=2 n \pi+\frac{3 \pi}{2} \
x=\frac{-\pi}{2} 	ext { are solution in }[-\pi, \pi] \
\cos \frac{3 x}{2}=0 \Rightarrow \frac{3 x}{2}=(2 n+1) \frac{\pi}{2} \
x=(2 n+1) \frac{\pi}{3} \quad \frac{\pi}{3}, \frac{-\pi}{3}, \pi,-\pi \
\cos \frac{x}{2}=0, \Rightarrow x=(2 n+1) \pi=\pi,-\pi
\end{gathered}
$$
All solutions $=x=\left\{-\pi, \frac{-\pi}{2}, \frac{-\pi}{3}, \frac{\pi}{3}, \piight\} 5$ solutions.