[tex]\dfrac{1}{\sqrt3}\cdot \sin(x+30)-\dfrac{\sqrt3}{2}\cdot \cos x=\\\\\dfrac{1}{\sqrt3}\cdot( \sin x\cos 30+\cos x\sin 30)-\dfrac{\sqrt3\cos x}{2}=\\\\\dfrac{1}{\sqrt3}\cdot\left( \sin x\cdot \dfrac{\sqrt3}{2}+\cos x\cdot\dfrac{1}{2}\right)-\dfrac{\sqrt3\cos x}{2}=\\\\\dfrac{\sin x}{2}+\dfrac{ \cos x}{2\sqrt3}-\dfrac{\sqrt3\cos x}{2}=\\\\\dfrac{\sin x-\sqrt3\cos x}{2}+\dfrac{\sqrt3 \cos x}{6}=\\\\\dfrac{3\sin x-3\sqrt3\cos x}{6}+\dfrac{\sqrt3 \cos x}{6}=[/tex]
[tex]\dfrac{3\sin x-3\sqrt3\cos x+\sqrt3 \cos x}{6}=\\\\\dfrac{3\sin x-2\sqrt3\cos x}{6}=\\\\\dfrac{3\sin x}{6}-\dfrac{2\sqrt3\cos x}{6}=\\\\\dfrac{\sin x}{2}-\dfrac{\sqrt3\cos x}{3}[/tex]
