Question

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

Answer 1

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) = - (1)/(2)`

`x = (2\pi)/(3) ,x = (4\pi)/(3)`

`\cos(x) - 1 = 0`

`\cos(x) = 1`

`x = 0`

So our answer are

`0, (2\pi)/(3) , (4\pi)/(3)`