Question

In the interval $[-\frac{\pi }{2},\frac{\pi }{2}]$, the equation ${\log }_{\sin \theta }\left(\cos 2\theta \right)=2$ has

• A. no solution
• B. a unique solution
• C. two solution
• D. infinitely many solution

Answer 1

The correct option is B a unique solution

We have, $-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$

$\therefore -1\le \sin \theta \le 1$, here $0<\sin \theta <1$.

Now, ${\log }_{\sin \theta }\left(\cos 2\theta \right)=2\Rightarrow \cos 2\theta ={\sin }^2\theta$

$\Rightarrow 1-2{\sin }^2\theta ={\sin }^2\theta \Rightarrow 3{\sin }^2\theta =1$

$\Rightarrow {\sin }^2\theta =\frac{1}{3}\Rightarrow \sin \theta =\frac{1}{\sqrt{3}}\{\because 0<\sin \theta <1\}$