# Question #a9636

Dec 14, 2016

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

$\implies \left(2 x + 1\right) = 2 n \pi - \left(2 x - 1\right) , \text{ where } n \in \mathbb{Z}$

$\implies 4 x = 2 n \pi$

$\implies x = \frac{1}{2} \left(n \pi\right)$

Dec 14, 2016

$x = \frac{k \pi}{2}$

#### Explanation:

Reminder:
cos x = cos a --> $x = \pm a + 2 k \pi$
Note: the equation can be written in radian form -->
$\cos \left(2 x + \frac{\pi}{3.14}\right) = \cos \left(2 x - \frac{\pi}{3.14}\right)$

Solve:
cos (2x + 1) = cos (2x - 1)
$2 x + 1 = \pm \left(2 x - 1\right) + 2 k \pi$

a. 2x + 1 = 2x - 1 + 2kpi
Equation insolvable

b. 2x + 1 = - 2x + 1 + 2kpi
$4 x = 2 k \pi$
$x = \frac{2 k \pi}{4} = \frac{k \pi}{2}$