Question

If $0<\theta <\frac{\pi }{2}$, then solution of the equation $\sin \theta -3\sin 2\theta +\sin 3\theta =\cos \theta -3\cos 2\theta +\cos 3\theta$ is

• A. $\frac{\pi }{16}$
• B. $\frac{\pi }{12}$
• C. $\frac{\pi }{8}$
• D. $\frac{\pi }{6}$

Answer 1

Answer: (C)

Given that,
$\sin \theta -3\sin 2\theta +\sin 3\theta =\cos \theta -3\cos 2\theta +\cos 3\theta$
$\Rightarrow (\sin \theta +\sin 3\theta )-3\sin 2\theta -(\cos \theta +\cos 3\theta )$
$+3\cos 2\theta =0$
$\Rightarrow 2\sin \left(\frac{\theta +3\theta }{2}\right)\cos \left(\frac{\theta -3\theta }{2}\right)-3\sin 2\theta$
$-2\cos \left(\frac{\theta +3\theta }{2}\right)\cos \left(\frac{\theta -3\theta }{2}\right)+3\cos 2\theta =0$
$\Rightarrow 2\sin 2\theta \cos \theta -3\sin 2\theta -2\cos 2\theta \cos \theta +3\cos 2\theta =0$
$\Rightarrow \sin 2\theta (2\cos \theta -3)-\cos 2\theta (2\cos \theta -3)=0$
$\Rightarrow (2\cos \theta -3)(\sin 2\theta -\cos 2\theta )=0$
$\Rightarrow 2\cos \theta -3=0\cos \theta =\frac{3}{2}$
which is not possible.
So, $\sin 2\theta -\cos 2\theta =0$
$\Rightarrow \sin 2\theta =\cos 2\theta$
$\Rightarrow \tan 2\theta =1$
$\Rightarrow \tan 2\theta =\tan \frac{\pi }{4}$
$\left[\because 0<\theta <\frac{\pi }{4}\right]$
$\Rightarrow 2\theta =\frac{\pi }{4}$
$\Rightarrow \theta =\frac{\pi }{8}$