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 $\cos$:

$\implies2\left(1-\dfrac{\sin\omega}{\cos\omega}\right)=\dfrac1{\cos^2\omega}$

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 $\cos2\omega$:

$\implies1=\dfrac{\sin2\omega}{\cos2\omega}=\tan2\omega$

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

$\omega=\dfrac\pi8,\dfrac{5\pi}8,\dfrac{9\pi}8,\dfrac{13\pi}8$