<s>[INST] Solve this trigonometric equation $ \sin2x-\sqrt3\cos2x=2$ Solve equation: $$ \sin2x-\sqrt3\cos2x=2$$
I tried dividing both sides with $\cos2x$ but then I win $\frac{2}{\cos2x}$. 
 [/INST] Notice, we have $$\sin 2x-\sqrt 3\cos 2x=2$$ $$\frac{1}{2}\sin 2x-\frac{\sqrt 3}{2}\cos 2x=1$$ $$\sin 2x\cos\frac{\pi}{3}-\cos 2x\sin \frac{\pi}{3}=1$$
$$\sin\left(2x-\frac{\pi}{3}\right)=1=\sin \frac{\pi}{2}$$
Now, writing the general solution for $x$, we get $$2x-\frac{\pi}{3}=2n\pi+\frac{\pi}{2}$$
$$2x=2n\pi+\frac{\pi}{2}+\frac{\pi}{3}=2n\pi+\frac{5\pi}{6}$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{x=n\pi+\frac{5\pi}{12}}}$$
Where, $n$ is any integer
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