A possible solution: First add 1 to both sides and reduce the RHS via the Pytagorean identity.
[tex]1-2\tan\omega=\tan^2\omega\implies2-2\tan\omega=1+\tan^2\omega=\sec^2\omega[/tex]
Rewrite [tex]\tan[/tex] and [tex]\sec[/tex] in terms of [tex]\sin[/tex] and [tex]\cos[/tex]:
[tex]\implies2\left(1-\dfrac{\sin\omega}{\cos\omega}\right)=\dfrac1{\cos^2\omega}[/tex]
Multiply both sides by [tex]\cos^2\theta[/tex]:
[tex]\implies2(\cos^2\omega-\sin\omega\cos\omega)=1[/tex]
[tex]\implies2\cos^2\omega-1=2\sin\omega\cos\omega[/tex]
Use the double angle identities:
[tex]\implies\cos2\omega=\sin2\omega[/tex]
Divide both sides by [tex]\cos2\omega[/tex]:
[tex]\implies1=\dfrac{\sin2\omega}{\cos2\omega}=\tan2\omega[/tex]
Now, [tex]\tan2\omega=1[/tex] for [tex]2\omega=\dfrac\pi4+n\pi[/tex], or [tex]\omega=\dfrac\pi8+\dfrac{n\pi}2[/tex], where [tex]n[/tex] is any integer. [tex]\omega[/tex] will fall in the interval [tex][0,2\pi)[/tex] for [tex]n=1,2,3,4[/tex], which means we have
[tex]\omega=\dfrac\pi8,\dfrac{5\pi}8,\dfrac{9\pi}8,\dfrac{13\pi}8[/tex]
