Question

Question #ef873

Answer 1

2 sets of solutions:

• `x = (2n+1)/2 pi`

• `x = 2npi`

For:
- `n in mathcal Z`

Explanation:

`sin^2 x + cos 2x - cos x =0`

Using identity: `cos 2 A = 1 - 2 sin^2 A`

`sin^2 x + 1 - 2 sin^2 x - cos x = 0`

`implies 1 - sin^2 x - cos x = 0`

Using identity: `sin^2 A + cos^2 A = 1`

`1 - (1 - cos^2 x) - cos x = 0`

`implies cos^2 x - cos x = 0`

`implies cos x (cos x - 1) = 0`

Looking at the 2 solutions:

• `cos x = 0 implies x = (2n+1)/2 pi`

• `cos x = 1 implies x = 2npi`