Question

Find all solutions in the region [0,2pi) for the equation cos^2(2x)-sin^2(2x)=0

Answer 1

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

`\rightarrow \sin^(2)(2x)=\cos^(2)(2x)`

`\rightarrow (\sin^(2)(2x))/(\cos^(2)(2x))=1`

`\rightarrow \tan^(2)(2x)=1`

`\rightarrow \tan(2x)=\pm 1`


`2x` can range anywhere in
`[0, 4\pi)`
So:

`\rightarrow 2x=(\pi)/(4), (3\pi)/(4), (5\pi)/(4), (7\pi)/(4), (9\pi)/(4), (11\pi)/(4), (13\pi)/(4), (15\pi)/(4)`

`\rightarrow x=(\pi)/(8), (3\pi)/(8), (5\pi)/(8), (7\pi)/(8), (9\pi)/(8), (11\pi)/(8), (13\pi)/(8), (15\pi)/(8)`