Tricky trigonometric problem

Question

I can not seem to solve this type of easy problem, it seems somewhat tricky, can I have a hint or an explanation.

The problem: $sin\left(2x\right)\:=\:\sqrt{3}cos\left(\:\frac{3\pi \:}{2}-x\right)$ on the interval $[ ; \frac{7}{2} ]$

Answer 1

The hint: $$\cos\left(\frac{3\pi}{2}-x\right)=-\sin{x}$$ and $$\sin2x=2\sin{x}\cos{x}.$$ Can you end it now?

Answer 2

Since $$\sin(2x)=2\sin(x)\cos(x)$$ and $$\cos(3/2\pi-x)=-\sin(x)$$ we get $$\sin(x)(2\cos(x)+\sqrt{3})=0$$ to solve.

Answer 3

$$ \sqrt{3}\cdot \cos(\frac{3\pi}{2}-x)\\ =\sqrt{3}\cdot \sin(\frac{3\pi}{2}+\frac{\pi}{2}-x)\\ = \sqrt{3}\cdot \sin(2\pi-x)\\ = \sqrt{3}\cdot \sin(-x)\\ = -\sqrt{3}\cdot \sin(x)\\ $$

$$ \sin 2x = 2\sin x\cos x $$

$$ \therefore,\\\ 2\sin x\cos x = -\sqrt{3}\cdot \sin(x)\\ $$ Now don't go cancelling stuff away first, acknowledge $\bf{\sin x=0}$ as a solution first. $$ 2\cos x = -\sqrt{3}\\ \bf{\cos x = -\frac{\sqrt{3}}{2}} $$ The two bolded parts give solutions and they are probably easy for you to work with.