Question

Consider the equation ${\log }_{\sqrt{2}\sin x}(1+\cos x)=2$, $x\in \left[-\frac{\pi }{2},\frac{3\pi }{2}\right]$. If the sum of the roots is $\frac{p\pi }{q}$, where G.C.D $(p,q)=1$, then the value of $p^2+q^2$ is

Answer 1

Answer: 10

${\log }_{\sqrt{2}\sin x}(1+\cos x)=2$
$\sqrt{2}\sin x\ne 1,\sqrt{2}\sin x>0,1+\cos x>0$
$\Rightarrow \sin x\ne \frac{1}{\sqrt{2}},\sin x>0$ and $x\ne 0$ or a multiple of $\pi$
$\Rightarrow x\in (0,\pi )-\left\{\frac{\pi }{4},\frac{3\pi }{4}\right\}$ (feasible region)
Hence, from (i), we get,
$(\sqrt{2}\sin x{)}^2=1+\cos x$
$\Rightarrow 2{\sin }^{2}x=1+\cos x$
$\Rightarrow 2{\cos }^{2}x+\cos x-1=0$
$\Rightarrow (2\cos x-1)(\cos x+1)=0$
$\Rightarrow \cos x=\frac{1}{2}[\cos x+1>0]$
$\Rightarrow x=\frac{\pi }{3}$
$\Rightarrow p=1,q=3$
$\Rightarrow p^2+q^2=10$