Question

Solve $4\sin \theta \sin 2\theta \sin 4\theta =\sin 3\theta$.

Answer 1

$4\sin \theta \sin 2\theta \sin 4\theta =\sin 3\theta$

$\Rightarrow 4\sin \theta \sin (3\theta -\theta )\sin (3\theta +\theta )=\sin 3\theta$

$\Rightarrow 4\left[\sin \theta \right({\sin }^{2}3\theta -{\sin }^2\theta \left)\right]=3\sin \theta -4{\sin }^3\theta$

$\Rightarrow 4\sin \theta {\sin }^{2}3\theta =3\sin \theta$

$\sin \theta (4{\sin }^{2}3\theta -3)=0$

$\Rightarrow \sin \theta =0$ or $4{\sin }^{2}3\theta -3=0$

$\Rightarrow \sin \theta =0$ or ${\sin }^{2}3\theta =\frac{3}{4}$

Now$\sin \theta =0\Rightarrow x=n\pi ,nϵZ$

And, ${\sin }^{2}3\theta =\frac{3}{4}$

$\Rightarrow {\sin }^{2}3\theta ={\left(\frac{\sqrt{3}}{2}\right)}^2$

$\Rightarrow {\sin }^{2}3\theta ={\sin }^2\frac{\pi }{3}$

$3\theta =m\pi \pm \frac{\pi }{3},mϵZ$

$x=\frac{m\pi }{3}\pm \frac{\pi }{9},mϵZ$

Hence $x=n\pi$ or $x=\frac{m\pi }{3}\pm \frac{\pi }{9}$ where

$m,nϵZ$