Question

The number of solutions of the equation $lo{g}_{\sqrt{2}sinx}(1+cosx)=2$ in the interval $\left[0,5\pi \right]$ is

Answer 1

Answer: 3

${\log }_{\sqrt{2}}\sin x(1+\cos x)=2$
$1+\cos x=(\sqrt{2}\sin x{)}^2=2{\sin }^{2}x$
$1+\cos x=2-2{\cos }^{2}x$
$2{\cos }^{2}x+\cos x-1=0$
$\Rightarrow (\cos x+1)(2\cos x-1)=0$
$\Rightarrow \cos x=\frac{1}{2}$ 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 $[0,5\pi ]$