# How do you solve trigonometric equations by factoring?

Nov 5, 2014

Example

Find $\theta$ in $\left[0 , 2 \pi\right)$ such that $2 {\sin}^{2} \theta + \sin \theta - 1 = 0$.

Since

$2 {x}^{2} + x - 1 = \left(2 x - 1\right) \left(x + 1\right)$,

by replacing $x$ by $\sin \theta$,

$2 {\sin}^{2} \theta + \sin \theta - 1 = \left(2 \sin \theta - 1\right) \left(\sin \theta + 1\right) = 0$

$\implies \left\{\left(\sin \theta = \frac{1}{2} \implies \theta = \frac{\pi}{6} \text{,} \frac{5 \pi}{6}\right) , \left(\sin \theta = - 1 \implies \theta = \frac{3 \pi}{2}\right)\right.$

Hence, the solutions are $\theta = \frac{\pi}{6} , \frac{5 \pi}{6} , \frac{3 \pi}{2}$.

I hope that this was helpful.