Question

The sum of all solutions of the equation $\cos 3\theta =\sin 2\theta$ in the interval $[-\frac{\pi }{2},\frac{\pi }{2}]$ is

• A. $-\frac{\pi }{5}$
• B. $-\frac{\pi }{3}$
• C. $\frac{\pi }{4}$
• D. $\frac{3\pi }{10}$

Answer 1

The correct option is A $-\frac{\pi }{5}$

Given equation can be written as
$\cos 3\theta =\cos (\frac{\pi }{2}-2\theta )\Rightarrow 3\theta =2n\pi \pm (\frac{\pi }{2}-2\theta )$
Now
$\Rightarrow 3\theta =2n\pi +(\frac{\pi }{2}-2\theta )\Rightarrow \theta =\frac{2n\pi +\frac{\pi }{2}}{5}$
and
$\Rightarrow 3\theta =2n\pi -(\frac{\pi }{2}-2\theta )\Rightarrow \theta =2n\pi -\frac{\pi }{2}$

Solution in $[-\frac{\pi }{2},\frac{\pi }{2}]$ are
$\{-\frac{\pi }{2},\frac{\pi }{10},+\frac{\pi }{2},-\frac{3\pi }{10}\}$
Hence sum of the solutions
$=-\frac{\pi }{2}+\frac{\pi }{10}+\frac{\pi }{2}-\frac{3\pi }{10}=-\frac{\pi }{5}$