Question

Solve ${\sin }^{2}n\theta -{\sin }^2(n-1)\theta ={\sin }^2\theta$

Answer 1

${\sin }^{2}n\theta -{\sin }^2(n+1)\theta ={\sin }^2\theta \Rightarrow \sin (n\theta +n\theta -\theta )\sin (n\theta -n\theta +\theta )={\sin }^2\theta \Rightarrow \sin (2n\theta -\theta )\sin \theta ={\sin }^2\theta \Rightarrow \sin \theta (\sin (2n\theta -\theta )-\sin \theta )=0$
$\Rightarrow \sin \theta \left(2\cos \left(\frac{2n\theta -\theta +\theta }{2}\right)\sin \left(\frac{2n\theta -\theta -\theta }{2}\right)\right)=0$
$\Rightarrow \sin \theta \left(2\cos n\theta \right)\sin (2n-1)\theta =0$
$\Rightarrow \sin \theta =0$ or $\cos n\theta =0$ or $\sin (n-1)\theta =0$
$\Rightarrow \theta =m\pi ,(2m+1)\frac{\pi }{2n},\frac{m\pi }{(n-1)}$