# Question 7142d

May 29, 2017

See below.

#### Explanation:

This is an identity:
$\cos 2 x = 2 {\cos}^{2} x - 1$

Plugging it in:
$2 \left(2 {\cos}^{2} - 1\right) = \cos x + 1$

$4 {\cos}^{2} - 2 = \cos x + 1$

$4 {\cos}^{2} - \cos x - 3 = 0$

If we set $\cos x = a$, then this is simply a quadratic, with two roots.

$4 {a}^{2} - a - 3 = 0$

$\left(4 a + 3\right) \left(a - 1\right) = 0$

$a = - \frac{3}{4} , 1$ or $\cos x = - \frac{3}{4} , 1$

These values all lie within the bounds of $\cos x$, so we can calculate them periodically,

$\cos x = 1$ when $x = 0 , 2 \pi , 4 \pi \ldots 2 n \pi$, or $0 , 360 , 720 , \ldots 360 n$, where $n$ is an integer.

$\cos x = - \frac{3}{4}$ is not as nice of a number.

${\cos}^{-} 1 \left(- \frac{3}{4}\right) = 2.41885841$ radians + 2nπ, or 138.59037789° + 360°n #