Question

Find all the values of $\theta$ satisfying the equation,$\sin \theta +\sin 5\theta =\sin 3\theta$ such that $0\le \theta \le \pi$

Answer 1

$\sin \theta +\sin 5\theta =\sin 3\theta$

$\Rightarrow 2\sin \left(\frac{\theta +5\theta }{2}\right)\cos \left(\frac{\theta -5\theta }{2}\right)=\sin 3\theta$

$\Rightarrow 2\sin \left(\frac{6\theta }{2}\right)\cos \left(\frac{-4\theta }{2}\right)=\sin 3\theta$

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

$\Rightarrow 2\sin 3\theta \cos \left(2\theta \right)=\sin 3\theta$

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

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

$\Rightarrow 3\theta =n\pi ,\cos 2\theta =\frac{1}{2}$

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

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

$\Rightarrow \theta =\frac{n\pi }{3},\theta =n\pi \pm \frac{\pi }{6}$

$\therefore \theta =0,\frac{\pi }{3},\frac{2\pi }{3},\frac{3\pi }{3}(=\pi )$

and $\theta =\frac{\pi }{6},\pi -\frac{\pi }{6}(=\frac{5\pi }{6})$

$\therefore$ the solutions are $\{0,\frac{\pi }{6},\frac{\pi }{3},\frac{2\pi }{3},\frac{5\pi }{6},\pi \}$