# How do you find all solutions of the equation in the interval (0, 2pi) 2cos^2 theta+cos theta = 0?

Apr 12, 2016

Factor out a cos.

#### Explanation:

$\cos \theta \left(2 \cos \theta + 1\right) = 0$

$\cos \theta = 0 \mathmr{and} 2 \cos \theta + 1 = 0$

$\theta = \pi , \frac{3 \pi}{2} \mathmr{and} \cos \theta = - \frac{1}{2}$

cos is negative in quadrants II and III. Applying the 30-60-90 special triangle and reference angles (leave a question on my homepage if you don't know how these work), we find that $\theta = \frac{2 \pi}{3} \mathmr{and} \frac{4 \pi}{3}$

Thus, your solution set is $\left\{\theta = \frac{3 \pi}{2} , \pi , \frac{2 \pi}{3} , \frac{4 \pi}{3}\right\}$

Hopefully this helps!