# a problem with condition $\sin(2\theta) = \tan(\theta) - \cos(2\theta)$

If $\theta$ represents an angle such that $\sin(2\theta) = \tan(\theta) - \cos(2\theta)$, then $\sin(\theta) - \cos(\theta)=$...?

I've been trying to do this problem for a while, but for some reason, I haven't been able to get the answer. I rewrote the $\sin(2\theta)$ as $2\sin(\theta)\cos(\theta)$ and $\tan(\theta) = \sin(\theta)/\cos(\theta)$ and $\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta)$ since there are both sines and cosines here, and I've tried a few other manipulations, but I don't know what to do from then on.

We have $$\sin(2\theta) = \tan(\theta) - \cos(2\theta)$$ so $$2\sin \theta\cos\theta=\frac{\sin\theta}{\cos\theta}-1+2\sin^2\theta$$ $$2\sin\theta\cos^2\theta=\sin\theta-\cos\theta+2\sin^2\theta\cos\theta$$ Hence $$\sin\theta-\cos\theta=2\sin\theta\cos\theta(\cos\theta-\sin\theta)=-\sin2\theta(\sin\theta-\cos\theta)$$giving $$(1+2\sin2\theta)(\sin\theta-\cos\theta)=0$$ So either $$\sin\theta-\cos\theta=0$$ or $$\sin2\theta=-\frac12\implies \theta=k\pi-\frac\pi{12}$$ for $k\in\mathbb{Z}$ thus $$\sin\theta-\cos\theta=\pm\frac{\sqrt6}2$$ (consider the cases when $k$ is even and odd)

Therefore $$\boxed{\sin\theta-\cos\theta=-\frac{\sqrt6}2, 0, \frac{\sqrt6}2}$$

Hint:

$t=\tan\theta,$

$$2t+1-t^2=t(1+t^2)\iff0=t^3+t^2-t-1=t^2(t+1)-(t+1)=(t+1)^2(t-1)$$

$\implies t=\pm1$

You have $$\sin2\theta-\tan\theta=-\cos2\theta$$ The left-hand side is $$2\sin\theta\cos\theta-\tan\theta=2\tan\theta\cos^2\theta-\tan\theta= \tan\theta(2\cos^2\theta-1)=\tan\theta\cos2\theta$$ so the equation becomes $$\cos2\theta(1+\tan\theta)=0$$ Can you finish?