<s>[INST] Tricky trigonometric problem 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 $[ ; \frac{7}{2}
]$
 [/INST] $$
\sqrt{3}\cdot \cos(\frac{3\pi}{2}-x)\\
=\sqrt{3}\cdot \sin(\frac{3\pi}{2}+\frac{\pi}{2}-x)\\
= \sqrt{3}\cdot \sin(2\pi-x)\\
= \sqrt{3}\cdot \sin(-x)\\
= -\sqrt{3}\cdot \sin(x)\\
$$
$$
\sin 2x = 2\sin x\cos x
$$
$$
\therefore,\\\
2\sin x\cos x = -\sqrt{3}\cdot \sin(x)\\
$$
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.
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