Question

If $\frac{k\pi }{10}$ is the least positive value of $\theta$ which satisfy the equation $\sin 3\theta +\cos 2\theta =0$, then $k$ is

Answer 1

$\sin 3\theta +\cos 2\theta =0$
$\Rightarrow -\sin 3\theta =\cos 2\theta$
$\Rightarrow \cos (\frac{\pi }{2}+3\theta )=\cos 2\theta$
$\Rightarrow \frac{\pi }{2}+3\theta =2n\pi \pm 2\theta ,n\in Z$

Case $\text{I}:$
$\frac{\pi }{2}+3\theta =2n\pi +2\theta$
$\Rightarrow \theta =2n\pi -\frac{\pi }{2}$
For the least positive value of $\theta$, put $n=1$
The least positive solution is $\theta =\frac{3\pi }{2}$

Case $\text{II}:$
$\frac{\pi }{2}+3\theta =2n\pi -2\theta$
$\Rightarrow 5\theta =2n\pi -\frac{\pi }{2}$
$\Rightarrow \theta =\frac{1}{5}(2n\pi -\frac{\pi }{2})$
For the least positive value of $\theta$, put $n=1$
The least positive solution is $\theta =\frac{3\pi }{10}$

Hence, the least positive value of $\theta$ which satisfies the equation is $\frac{3\pi }{10}$