Question

Find all exact solutions that exist on the interval [0, 2π).

1 − 2 tan(ω) = tan2(ω)

w =

Answer 1

A possible solution: First add 1 to both sides and reduce the RHS via the Pytagorean identity.

`1-2\tan\omega=\tan^2\omega\implies2-2\tan\omega=1+\tan^2\omega=\sec^2\omega`

Rewrite
`\tan` and
`\sec` in terms of
`\sin` and
`\cos`:

`\implies2\left(1-(\sin\omega)/(\cos\omega)\right)=\frac1{\cos^2\omega}`

Multiply both sides by
`\cos^2\theta`:

`\implies2(\cos^2\omega-\sin\omega\cos\omega)=1`

`\implies2\cos^2\omega-1=2\sin\omega\cos\omega`

Use the double angle identities:

`\implies\cos2\omega=\sin2\omega`

Divide both sides by
`\cos2\omega`:

`\implies1=(\sin2\omega)/(\cos2\omega)=\tan2\omega`

Now,
`\tan2\omega=1` for
`2\omega=\frac\pi4+n\pi`, or
`\omega=\frac\pi8+\frac{n\pi}2`, where
`n` is any integer.
`\omega` will fall in the interval
`[0,2\pi)` for
`n=1,2,3,4`, which means we have

`\omega=\frac\pi8,\frac{5\pi}8,\frac{9\pi}8,\frac{13\pi}8`