Question

The general value of $\theta$in the equation$2\sqrt{3}\cos\theta = \tan\theta,$ is

• A. $2n\pi \pm \frac{\pi}{6}$
• B. $2n\pi \pm \frac{\pi}{4}$
• C. $n\pi + ( - 1)^{n}\frac{\pi}{3}$
• D. $n\pi + ( - 1)^{n}\frac{\pi}{4}$

Answer 1

Answer: (C)

$n\pi + ( - 1)^{n}\frac{\pi}{3}$

Answer 2

$2\sqrt{3}\cos^{2}\theta = \sin\theta \Rightarrow 2\sqrt{3}\sin^{2}\theta + \sin\theta - 2\sqrt{3} = 0$

$\mathbf{\Rightarrow}\mathbf{\sin}\mathbf{\theta}\mathbf{=}\frac{\mathbf{- 1 \pm 7}}{\mathbf{4}\sqrt{\mathbf{3}}}\mathbf{\Rightarrow}\mathbf{\sin}\mathbf{\theta}\mathbf{=}\frac{\mathbf{-}\mathbf{8}}{\mathbf{4}\sqrt{\mathbf{3}}}$ (impossible) or

$\sin\theta = \frac{6}{4\sqrt{3}} = \frac{\sqrt{3}}{2}$ $\Rightarrow \theta = n\pi + ( - 1)^{n}\frac{\pi}{3}.$