# How do you solve the equation 2cos^2theta-3costheta+1=0 for 0<=theta<2pi?

Nov 21, 2016

The solutions are $S = \left\{0 , \frac{\pi}{3} , \frac{5 \pi}{3}\right\}$

#### Explanation:

Let's factorise the LHS

$2 {\cos}^{2} \theta - 3 \cos \theta + 1 = \left(2 \cos \theta - 1\right) \left(\cos \theta - 1\right) = 0$

Therefore, $2 \cos \theta - 1 = 0$

$\cos \theta = \frac{1}{2}$

and $\cos \theta - 1 = 0$

$\cos \theta = 1$

We are looking for theta in [0, 2pi[

When $\cos \theta = \frac{1}{2}$

$\theta = \frac{\pi}{3} , \frac{5 \pi}{3}$

When $\cos \theta = 1$

$\theta = 0$