# How do you solve 2cos^2theta-4costheta-5=0 in the interval 0<=x<=2pi?

Oct 3, 2016

$\theta = 150.56$ deg

#### Explanation:

$2 {\cos}^{2} \theta - 4 \cos \theta - 5 = 0$. Note that this equation looks like the quadratic equation $2 {x}^{2} - 4 x - 5 = 0$. So we will solve using the quadratic formula $x = \frac{- b \pm \sqrt{\left({b}^{2} - 4 a c\right)}}{2 a}$
$a = 2 , b = - 4 , c = - 5$

$\cos \theta = \frac{4 \pm \sqrt{16 - 4 \left(2\right) \left(- 5\right)}}{4}$

$\cos \theta = \frac{4 \pm \sqrt{56}}{4}$

$\cos \theta = \frac{4 + \sqrt{56}}{4}$ or $\cos \theta = \frac{4 - \sqrt{56}}{4}$

$\theta = {\cos}^{-} 1 \left(\frac{4 + \sqrt{56}}{4}\right)$ or $\theta = {\cos}^{-} 1 \left(\frac{4 - \sqrt{56}}{4}\right)$
$\theta =$undefined or $\theta = 150.56$ deg