Question

What is the answer of
2sin^2(x)-cos(2x)=0 ?

Answer 1

`2\sin^2x-\cos2x=2\sin^2x-(2\cos^2x-1)=2(\sin^2-\cos^2x)+1=0`

`-2\cos2x=-1`

`\cos2x=\frac12`

You have
`\cos x=\frac12` for


`x=\frac\pi3+2n\pi`

`x=-\frac\pi3+2n\pi`

which means
`\cos 2x=\frac12` for


`2x=\frac\pi3+2n\pi\implies x=\frac\pi6+n\pi`

`2x=-\frac\pi3+2n\pi\implies x=-\frac\pi6+n\pi`

where
`n` is any integer.