Question

Numbers of solution of the equation $\sin 7\theta =\sin \theta +\sin 3\theta for\theta \in \left(0,\pi \right)$ is equal to

Answer 1

$\sin 7\theta -\sin \theta -\sin 3\theta =0$

$\Rightarrow 2\cos \left(\frac{7\theta +\theta }{2}\right)\sin \left(\frac{7\theta -\theta }{2}\right)-\sin 3\theta =0$

$\Rightarrow 2\cos \left(\frac{8\theta }{2}\right)\sin \left(\frac{6\theta }{2}\right)-\sin 3\theta =0$

$\Rightarrow 2\cos 4\theta \sin 3\theta -\sin 3\theta =0$

$\Rightarrow \sin 3\theta (2\cos 4\theta -1)=0$

$\Rightarrow \sin 3\theta =0,2\cos 4\theta =1$

$\Rightarrow \sin 3\theta =0,\cos 4\theta =\frac{1}{2}$

$\Rightarrow 3\theta =n\pi ,4\theta =2n\pi \pm \frac{\pi }{3}$

$\Rightarrow \theta =\frac{n\pi }{3},\theta =\frac{n\pi }{2}\pm \frac{\pi }{12}$

$\therefore \theta =\{\frac{\pi }{3},\frac{2\pi }{3}\}\cup \{\frac{\pi }{12},\frac{\pi }{2}+\frac{\pi }{12},\frac{\pi }{2}-\frac{\pi }{12},\pi -\frac{\pi }{12}\}$

$=\{\frac{\pi }{3},\frac{2\pi }{3},\frac{\pi }{12},\frac{6\pi +\pi }{12},\frac{6\pi -\pi }{12},\frac{11\pi }{12}\}$

$=\{\frac{\pi }{3},\frac{2\pi }{3},\frac{\pi }{12},\frac{7\pi }{12},\frac{5\pi }{12},\frac{11\pi }{12}\}$

$=\{\frac{\pi }{12},\frac{\pi }{3},\frac{7\pi }{12},\frac{5\pi }{12},\frac{2\pi }{3},\frac{11\pi }{12}\}$

$\therefore$Number of solutions$=6$