Question

Solve the following equation:

\[3 \cos^2 x - 2\sqrt{3} \sin x \cos x - 3 \sin^2 x = 0\]

Answer 1

\[3 \cos^2 x - 2\sqrt{3} \sin x \cos x - 3 \sin^2 x = 0\]

Now,

\[3 ( \cos^2 x - \sin^2 x) - \sqrt{3} \sin2x = 0\]

\[ \Rightarrow 3 \cos2x - \sqrt{3} \sin2x = 0\]

\[ \Rightarrow \sqrt{3} (\sqrt{3} \cos2x - \sin2x) = 0\]

\[ \Rightarrow (\sqrt{3} \cos2x - \sin2x) = 0\]

\[ \Rightarrow \frac{\sin2x}{\cos2x} = \sqrt{3} \]

\[ \Rightarrow \tan2x = \tan \frac{\pi}{3}\]

\[ \Rightarrow 2x = n\pi + \frac{\pi}{3}, n \in Z\]

\[ \Rightarrow x = \frac{n\pi}{2} + \frac{\pi}{6}, n \in Z\]