# How do you solve sqrt2 sin2theta+1=0?

Jun 24, 2018

$\theta = \frac{5 \pi}{8} , \frac{7 \pi}{8} , \frac{13 \pi}{8} , \frac{15 \pi}{8}$ if the domain is $0 \le \theta \le 2 \pi$

#### Explanation:

$\sqrt{2} \sin 2 \theta + 1 = 0$

$\sqrt{2} \sin 2 \theta = - 1$

$\sin 2 \theta = - \frac{1}{\sqrt{2}}$

Remember that $\sin 2 \theta$ is negative in the 3rd and 4th quadrants

$2 \theta = \pi + \frac{\pi}{4} , 2 \pi - \frac{\pi}{4} , 3 \pi + \frac{\pi}{4} , 4 \pi - \frac{\pi}{4}$

$2 \theta = \frac{5 \pi}{4} , \frac{7 \pi}{4} , \frac{13 \pi}{4} , \frac{15 \pi}{4}$

$\theta = \frac{5 \pi}{8} , \frac{7 \pi}{8} , \frac{13 \pi}{8} , \frac{15 \pi}{8}$

if the domain is $0 \le \theta \le 2 \pi$