Question

Solve ${\sin }^3\theta \cos \theta -{\cos }^3\theta \sin \theta =1/4$.

Answer 1

We write the given equation as
$4\sin \theta \cos \theta ({\sin }^2\theta -{\cos }^2\theta )=1$
or $2\sin 2\theta (-\cos 2\theta )=1$ or $-\sin 4\theta =1$
or $\sin 4\theta =-1$
$\therefore 4\theta =2n\pi +\left(3\pi /2\right)$
or $\theta =\left(n\pi /2\right)+\left(3\pi /8\right),n\in I$.