<s>[INST] $\ 3\,\tan(x)-\sqrt3 $ You have to find alternate form, that has only multiplication/division signs between trigonometric functions 
The form is: $\ 3\,\tan(x)-\sqrt3 $  
The solution form is: $\ \left(\frac{2*\sqrt 3\sin(x-π/6)}{\cos(x)}\right) $
My try: $\ 3\,\tan(x)-\sqrt3 = 3*\left(\frac{\sin(x)}{\cos(x)}\right)-\left(\frac{\sin(π/3)}{\cos(π/3)}\right)= \left(\frac{\sin(x)\cos(π/3)-\sin(π/3)\cos(x)}{\cos(x)\cos(π/3)}\right)=\left(\frac{\sin(x)\cos(π/3)+\sin(x)\cos(π/3)-\sin(π/(3)\cos(x)}{\cos(x)\cos(π/3)}\right)$
I am certain that I have done something wrong in one of the steps, as this leads to only more complicated form, that isn't even near the solution
 [/INST] Note that we can write $$3\tan x- \sqrt{3} = 2\times \frac{\sqrt{3}}{2}\frac{\sin x}{\cos x}\times \sqrt{3} - 2\times \sqrt{3}\times \frac{\cos x}{\cos x}\frac{1}{2}$$ $$=2\times \sqrt{3}\frac{1}{\cos x}[\sin x(\frac{\sqrt{3}}{2})-\cos x(\frac{1}{2})]$$ $$=\frac{2\times \sqrt{3}}{\cos x}[\sin x\cos \frac{\pi}{6}-\cos x\frac{\pi}{6}]$$ $$=\frac{2\sqrt{3}\sin(x-\frac{\pi}{6})}{\cos x}$$
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