How do you solve 2sin^2theta=1?

Nov 20, 2016

$\theta = \frac{\left(2 k + 1\right)}{4} \pi , k \in m a t h \boldsymbol{Z}$

Explanation:

$2 {\sin}^{2} \theta = 1$
${\sin}^{2} \theta = \frac{1}{2}$
$\sin \theta = \sqrt{\frac{1}{2}} \mathmr{and} \sin \theta = - \sqrt{\frac{1}{2}}$
$\sin \theta = \frac{\sqrt{2}}{2} \mathmr{and} \sin \theta = - \frac{\sqrt{2}}{2}$
$\theta = \frac{\pi}{4} + 2 k \pi \mathmr{and} \theta = \pi - \frac{\pi}{4} + 2 k \pi \mathmr{and} \theta = - \frac{\pi}{4} + 2 k \pi \mathmr{and} \theta = - \left(\pi - \frac{\pi}{4}\right) + 2 k \pi , k \in m a t h \boldsymbol{Z}$
$\theta = \frac{\pi}{4} + k \pi \mathmr{and} \theta = - \frac{\pi}{4} + k \pi , k \in m a t h \boldsymbol{Z}$
$\theta = \frac{\pi}{4} + k \frac{\pi}{2} , k \in m a t h \boldsymbol{Z}$
or written a bit differently
$\theta = \frac{\left(2 k + 1\right)}{4} \pi , k \in m a t h \boldsymbol{Z}$