Question

Sin^2Ф-cos^2Ф=0

The interval is 0≤Ф<2
`\pi`

Answer 1

Recall the double angle identity for cosine:

`\cos2\Phi=\cos^2\Phi-\sin^2\Phi`

So we have

`\sin^2\Phi-\cos^2\Phi=-\cos2\Phi=0`

`\cos x=0` for
`x=\frac{(2n+1)\pi}2`, where
`n` is any integer, so

`2\Phi=\frac{(2n+1)\pi}2\implies\Phi=\frac{(2n+1)\pi}4`

for any integer
`n`. We get solutions in the interval
`0\le\Phi<2\pi` when
`n=1,2,3,4`, giving

`\Phi=\frac\pi4,\frac{3\pi}4,\frac{5\pi}4,\frac{7\pi}4`