Question

How do you solve the identity `2 sin^2 x = 2 + cos x`?

Answer 1

`x = (2n-1)\frac{\pi}{2} " or " x = (2n-1)\pi \pm {\pi}/3 , " " n inZZ`

Explanation:

`sin^2 x + cos^2 x -= 1`
`\therefore 2sin^2 x = 2-2cos^2 x`

Substituting above to original equation:
`2-2cos^2 x = 2 + cos x`

Re-arrange:
`2cos^2 x + cos x = 0`
`cos x(2cos x + 1) = 0`

`cos x = 0 " or " cos x = -\frac{1}{2}`

`x = (2n-1)\frac{\pi}{2} " or " x = (2n-1)\pi \pm {\pi}/3 , " " n in ZZ`