سؤال #6,004

$$3\cos^{2} x - \sqrt{3}\sin x\cos x = 3 \Rightarrow 3\cos^{2} x - 3 - \sqrt{3}\sin x\cos x = 0 \Rightarrow $$ $$3(\cos^{2} x - 1) - \sqrt{3}\sin x\cos x = 0 \Rightarrow -3\sin^{2} x - \sqrt{3}\sin x\cos x = 0 \Rightarrow $$ $$\sin x(-3\sin x - \sqrt{3}\cos x) = 0$$ $$\begin{cases} \sin & x = 0 \Rightarrow x = k\pi \\ -3\sin & x -\sqrt{3}\cos x= 0\Rightarrow -3\sin x= \sqrt{3}\cos x\Rightarrow \mathrm{tg} x= -\frac{\sqrt{3}}{3}= \mathrm{tg}(-\frac{\pi}{6})\Rightarrow x= k\pi - \frac{\pi}{6} \end{cases}$$ بنابراین گزینه 2 صحیح است.