# How do you solve \sin^2 \theta = 2 \sin^2 \frac{\theta}{2}  over the interval [0,2pi]?

Nov 20, 2014

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

by ${\sin}^{2} \theta = 1 - {\cos}^{2} \theta$ and ${\sin}^{2} \left(\frac{\theta}{2}\right) = \frac{1}{2} \left(1 - \cos \theta\right)$,

$\implies 1 - {\cos}^{2} \theta = 1 - \cos \theta$

by subtracting $1$,

$\implies - {\cos}^{2} \theta = - \cos \theta$

by adding $\cos \theta$,

$\implies \cos \theta - {\cos}^{2} \theta = 0$

by factoring out $\cos \theta$,

$\implies \cos \theta \left(1 - \cos \theta\right) = 0$

$\implies \left\{\begin{matrix}\cos \theta = 0 \implies \theta = \frac{\pi}{2} \text{ & "{3pi}/2 \\ cos theta=1 => theta=0" & } 2 \pi\end{matrix}\right.$

Hence, $\theta = 0 , \frac{\pi}{2} , \frac{3 \pi}{2} , 2 \pi$.

I hope that this was helpful.