Question

The number of solutions of the equation $log_{\sqrt{2} sin x}\left(\right.1+cos⁡x\left.\right)=2$ in the interval $\left[0 , 5 \pi \right]$ is

Answer 1

$\log _{\sqrt{2}} \sin x(1+\cos x)=2$
$1+\cos x=(\sqrt{2} \sin x)^{2}=2 \sin ^{2} x$
$1+\cos x=2-2 \cos ^{2} x$
$2 \cos ^{2} x+\cos x-1=0$
$\Rightarrow (\cos x+1)(2 \cos x-1)=0$
$\Rightarrow \cos x=\frac{1}{2}$ or $\cos x=-1$ (rejected)
$\Rightarrow x=\frac{\pi}{3}, \frac{5 \pi}{3}$ in $[0,2 \pi]$
But at $x=\frac{5 \pi}{3}, \sqrt{2} \sin x$ is negative
$\Rightarrow x=\frac{\pi}{3}$ in $[0,2 \pi]$
$\Rightarrow x=\frac{\pi}{3}, \frac{\pi}{3}+2 \pi, \frac{\pi}{3}+4 \pi$
$\Rightarrow 3$ solutions in $[0,5 \pi]$