<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] $\bf{My\; Solution::}$ Given $\displaystyle \sin 2x -\sqrt{3}\cos 2x = 2$
We can write it as $$\displaystyle \sin 2x \cdot \frac{1}{2}-\cos 2x\cdot  \frac{\sqrt{3}}{2} = 1\Rightarrow \sin \left(2x-\frac{\pi}{3}\right) =1=\sin \frac{\pi}{2}$$
Above we have used the formula $$ \sin \alpha\cdot \cos \beta-\cos \alpha\cdot \sin \beta=\sin (\alpha-\beta).$$
So $$\displaystyle 2x-\frac{\pi}{3} = n\pi+(-1)^n\cdot \frac{\pi}{2}\Rightarrow x=\frac{n\pi}{2}+(-1)^n\cdot \frac{\pi}{4}+\frac{\pi}{6}\;,$$ Where $n\in \mathbb{Z}$
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