Solve : sin^2nƟ - sin^2(n-1) Ɵ = sin^2 Ɵ
I can't tell whether your equation is $\displaystyle \displaystyle \begin{align*} \sin^{2n}{\theta} - \sin^{2(n-1)}{\theta} = \sin^2{\theta} \end{align*}$ or $\displaystyle \displaystyle \begin{align*} \sin^2{(n\theta)} - \sin^2{[(n-1)\theta]} = \sin^2{\theta} \end{align*}$...
the equation is not an identity, however (after playing with the calculator) ...
for $\displaystyle n = 2$
$\displaystyle \sin^2(2\theta) - \sin^2{\theta} = \sin^2{\theta}$ for $\displaystyle \theta$ equal to integral multiples of $\displaystyle \pi$ and odd multiples of $\displaystyle \frac{\pi}{4}$
for $\displaystyle n = 3$
$\displaystyle \sin^2(3\theta) - \sin^2{2\theta} = \sin^2{\theta}$ for $\displaystyle \theta$ equal to integral multiples of $\displaystyle \pi$ and odd multiples of $\displaystyle \frac{\pi}{6}$
for $\displaystyle n = 4$
$\displaystyle \sin^2(4\theta) - \sin^2{3\theta} = \sin^2{\theta}$ for $\displaystyle \theta$ equal to integral multiples of $\displaystyle \pi$ and odd multiples of $\displaystyle \frac{\pi}{8}$
hmmm ...
conjecture that $\displaystyle \sin^2(n\theta) - \sin^2[(n-1)\theta] = \sin^2{\theta}$ for $\displaystyle \theta$ equal to integral multiples of $\displaystyle \pi$ and odd multiples of $\displaystyle \frac{\pi}{2n}$
... ???
It needs the use of three trig identities.
$\displaystyle \sin A + \sin B=2\sin \left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)..................(1)$
$\displaystyle \sin A - \sin B=2\cos \left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)..................(2)$
$\displaystyle \sin 2A = 2\sin A \cos A................................................. .......(3)$
Briefly, take a difference of two squares on the LHS, simplify the two brackets using (1) and (2) and then use (3) (twice) to simplify the resulting expression. Move the resulting expression to the RHS, remove $\displaystyle \sin \theta$ as a common factor and then use (2) again on the expression within the brackets.
That leads to
$\displaystyle 2\sin \theta \cos n\theta \sin(n-1)\theta =0$ from which $\displaystyle \sin\theta=0,$ or $\displaystyle \cos n\theta=0$ or $\displaystyle \sin (n-1)\theta = 0.$