Question

Number of solutions of the equation $2({\sin }^3\theta +{\sin }^2\theta )+2({\cos }^3\theta +{\cos }^2\theta )=3\sin 2\theta$
in the interval $[0,4\pi ]$ is

Answer 1

$2({\sin }^3\theta +{\sin }^2\theta )+2({\cos }^3\theta +{\cos }^2\theta )=3\sin 2\theta \Rightarrow {\sin }^3\theta +{\cos }^3\theta +1=3\sin \theta \cos \theta \Rightarrow \sin \theta +\cos \theta +1=0(\because a^3+b^3+c^3-3abc=(a+b+c\left)\right(a^2+b^2+c^2-ab-bc-ca\left)\right)$