# How do you solve the equation 2cos^2x- 7cosx + 3 = 0?

Aug 30, 2016

$x = \frac{\pi}{3} + 2 \pi n \mathmr{and} x = \frac{5 \pi}{3} + 2 \pi n$

#### Explanation:

Solve by factoring.

$2 {\cos}^{2} x - 6 \cos x - \cos x + 3 = 0$

$2 \cos x \left(\cos x - 3\right) - 1 \left(\cos x - 3\right) = 0$

$\left(2 \cos x - 1\right) \left(\cos x - 3\right) = 0$

$\cos x = \frac{1}{2} \mathmr{and} \cos x = 3$

$x = \frac{\pi}{3} + 2 \pi n , x = \frac{5 \pi}{3} + 2 \pi n$

Hopefully this helps!