Question

How do you solve `cos2x + sin^2x = 0` from 0 to 2pi?

Answer 1

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

Explanation:

One form of the double-angle formula for cosine is `cos(2x)=1-2sin^{2}(x)` (this is not an equation to solve, it's an "identity", meaning it's true for all `x` where it's defined, which is for all `x\in RR`).

Using this identity, we can re-write `cos(2x)+sin^{2}(x)=0` as `1-2sin^{2}(x)+sin^{2}(x)=0`, or `1-sin^{2}(x)=0`, or `sin^{2}(x)=1`.

The only values of `x` where `sin^{2](x)=1` on the interval `[0,2pi]` are `x=pi/2` and `x=(3pi)/2`.