Question

Solve the trigonometric equation ${\tan }^2\theta +(1-\sqrt{3})\tan \theta -\sqrt{3}=0$

Answer 1

${\tan }^2\theta +(1-\sqrt{3})\tan \theta -\sqrt{3}=0$

$\tan \theta =\frac{(\sqrt{3}-1)\pm \sqrt{(1-\sqrt{3}{)}^2+4\sqrt{3}}}{2}$

$=\frac{(\sqrt{3}-1)\pm (1+\sqrt{3})}{2}$

$=\sqrt{3}$ or $-1$

$\therefore \theta =n\pi +\frac{\pi }{3}$ or $n\pi -\frac{\pi }{4}$