Question

Question #97e1b

Answer 1

`x=pi*n` where `n in ZZ`.

Explanation:

The Pythagorean Identity says: `cos^2(x)+sin^2(x)=1` which means we can rewrite the original:

`2sin^2(x)=1-1 rarr 2sin^2(x)=0 rarr sin(x)=0`

If `sin(x)=0` we know `x=0+2pi*n` where `n in ZZ` or `x= pi + 2pi*n` where `n in ZZ`.

We could also simplify this to `x=pi*n` where `n in ZZ`.