Question

# Solve the following trigonometric equation:$3{\cos ^2}\theta - 2\sqrt 3 \cos \theta \sin \theta - 3{\sin ^2}\theta = 0$.

Hint: By the use of trigonometric formulae we need to find the value of $'\theta '$ satisfying the given equation.
$3{\cos ^2}\theta - 2\sqrt 3 \cos \theta \sin \theta - 3{\sin ^2}\theta = 0 \to (1)$
$\Rightarrow 3({\cos ^2}\theta - {\sin ^2}\theta ) - \sqrt 3 (2\cos \theta \sin \theta ) = 0 \\ \Rightarrow 3({\cos ^2}\theta - {\sin ^2}\theta ) = \sqrt 3 (2\cos \theta \sin \theta ) \to (2) \\$
We know that $\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta$ and $\sin 2\theta = 2\sin \theta \cos \theta$. Substituting in equation (2), we get
$\Rightarrow 3\cos 2\theta = \sqrt 3 \sin 2\theta \\ \Rightarrow \dfrac{3}{{\sqrt 3 }} = \dfrac{{\sin 2\theta }}{{\cos 2\theta }} \\ \Rightarrow \tan 2\theta = \sqrt 3 \\$
Now, let us write the value of $\sqrt 3$ in terms of $\tan$ i.e.., $\tan (\dfrac{\pi }{3}) = \sqrt 3$, Substituting in the above equation, we get
$\Rightarrow \tan 2\theta = \tan (\dfrac{\pi }{3}) \\ \Rightarrow 2\theta = n\pi + \dfrac{\pi }{3} \\ \Rightarrow \theta = \dfrac{{n\pi }}{2} + \dfrac{\pi }{6} \\$
Hence, the value of $'\theta '$ satisfying the given equation is $\dfrac{{n\pi }}{2} + \dfrac{\pi }{6}$, where n is an integer.
Note: Here, we have added $'n\pi '$ to the $\dfrac{\pi }{3}$ after cancelling the tan on the both sides as $'\pi '$ is the period of the tan function and n is an integral number.