Question

Question #d62d3

Answer 1

`cos(2x) = 1-2sin^2x`
`2sinx cos2x = 0`
`2sinx(1-2sin^2x) = 0`
Let `y = sinx`

`2y(1-2y^2) = 0`

Then we have
`y = 0` or `1-2y^2 = 0`

For `y=0`
`sinx = 0`
`x = n\pi` where `n\in{1,2,3,....}`

For `1-2y^2 = 0`
`y = \pm1/sqrt(2)`
`sinx = +- 1/sqrt(2)`
`x = (n\pi)/4` where `n \in {1,2,3,....)`