Question

Question #8d8dd

Answer 1

`x = 0, pi`

Explanation:

`2cos ^2 x- 2 = 0`
`2cos^2x= 2`
`cos^2x = 1`
`cos x = +-1`
`x = 0, pi`

Aliter :
`2cos^2x -2 = 0`
`2cos^2x -1 = 1`
`cos (2x) = 1`
`2x = cos^(-1)1 = +-0`
`x = 0, pi`