# How do you solve 2cos^2 theta + cos theta -1 = 0?

Sep 15, 2016

(1 + 2k)pi; pi/3 + 2kpi; (5pi)/3 + 2kpi

#### Explanation:

Solve the quadratic equation for cos t.
$f \left(t\right) = 2 {\cos}^{2} t + \cos t - 1 = 0$
Since a - b + c = 0, use shortcut. The 2 real roots are:
cos t = -1 and $\cos t = - \frac{c}{a} = \frac{1}{2}$
Use trig table of special arcs and unit circle -->

a. cos t = -1 --> $t = \pi + 2 k \pi$
General answers: $t = \left(1 + 2 k\right) \pi$
b. $\cos t = \frac{1}{2}$ -->$t = \pm \frac{\pi}{3} + 2 k \pi$
arc $\frac{- \pi}{3}$ and arc $\frac{5 \pi}{3}$ are co-terminal.
$\left(1 + 2 k\right) \pi$
$\frac{\pi}{3} + 2 k \pi$
$\frac{5 \pi}{3} + 2 k \pi$