Question

General solution of $\tan\theta = \frac{1}{2}\left( \sqrt{3} - \frac{1}{\sqrt{3}} \right) = \frac{1}{\sqrt{3}} = \tan\left( \frac{\pi}{6} \right)$ is.

• A. $\Rightarrow$
• B. $\theta = n\pi + \frac{\pi}{6}$
• C. $\therefore$
• D. $0 < \theta < 2\pi$

Answer 1

Answer: (A)

$\Rightarrow$

Answer 2

$\tan 5 \theta = \tan \left( \frac { \pi } { 2 } - 2 \theta \right)$ $\sin 2\theta = \cos 3\theta$ $\Rightarrow$

$3\theta = 2n\pi \pm \left( \frac{\pi}{2} - 2\theta \right)$ $\Rightarrow \theta = \frac{2n\pi}{5} + \frac{\pi}{10}$ $\theta = 2n\pi - \frac{\pi}{2}$.