Question

If $2{\sin }^2\theta +\sqrt{3}\cos \theta +1=0$, then the value of $\theta$ is

• A. $\frac{\pi }{6}$
• B. $\frac{2\pi }{3}$
• C. $\frac{5\pi }{6}$
• D. $\pi$

Answer 1

Answer: (C)

$\frac{5\pi }{6}$

Given, $2{\sin }^2\theta +\sqrt{3}\cos \theta +1=0$
$\Rightarrow 2(1-{\cos }^2\theta )+\sqrt{3}\cos \theta +1=0$
$\Rightarrow 2{\cos }^2\theta -\sqrt{3}\cos \theta -3=0$
$\therefore \cos \theta =\frac{\sqrt{3}\pm \sqrt{3+4\times 3\times 2}}{2\times 2}$
$=\frac{\sqrt{3}\pm 3\sqrt{3}}{4}$
$\Rightarrow \cos \theta =-\frac{\sqrt{3}}{2}$
and $\cos \theta =\sqrt{3}$
$\therefore \cos \theta =-\frac{\sqrt{3}}{2}$
$(\because \cos \theta$ cannot be greater than $1$)
$\Rightarrow \theta =\frac{5\pi }{6}$.