Question

Find all solutions to the equation in the interval [0,2pi). Enter the solutions in increasing order. Cos 2x = cos x

Answer 1

Step-by-step explanation:

$\cos(2x) = \cos(x)$

$\cos {}^{2} (x) - \sin {}^{2} (x) = \cos(x)$

$\cos {}^{2} (x) - (1 - \cos {}^{2} (x) ) = \cos(x)$

$2 \cos {}^{2} (x) - 1 = \cos(x)$

$2 \cos {}^{2} (x) - \cos(x) - 1 = 0$

$2 \cos {}^{2} (x) - 2 \cos(x) + \cos(x) - 1 = 0$

$2 \cos(x) ( \cos(x) - 1) + 1( \cos(x) + 1)$

$(2 \cos(x) + 1)( \cos(x) - 1) = 0$

$2 \cos(x) + 1 = 0$

$2 \cos(x) = - 1$

$\cos(x) = - \frac{1}{2}$

$x = \frac{2\pi}{3} ,x = \frac{4\pi}{3}$

$\cos(x) - 1 = 0$

$\cos(x) = 1$

$x = 0$

So our answer are

$0, \frac{2\pi}{3} , \frac{4\pi}{3}$