Question

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

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

Answer 1

The correct option is C

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

Explanation for the correct option:

Step 1. Find the value of $\theta$:

Given,

$2{\sin }^2\theta +\sqrt{3}\cos \theta +1=0$

$\Rightarrow$$2(1–{\cos }^2\theta )+\sqrt{3}\cos \theta +1=0$ $\left[\mathbf{\because }\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\theta }\mathbf{}\mathbf{+}\mathbf{}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\right]$

$\Rightarrow$ $2–2{\cos }^2\theta +\sqrt{3}\cos \theta +1=0$

$\Rightarrow$ $2{\cos }^2\theta –\sqrt{3}\cos \theta –3=0$

It makes a quadratic equation in $\cos$ θ

Step 2. Apply quadratic formula:

$\begin{array}{rcl}\cos \theta & =& \frac{[\sqrt{3}\pm \sqrt{(3+24)}]}{4}\\ & =& \frac{(\sqrt{3}\pm 3\sqrt{3})}{4}\end{array}$ $\left[\mathbf{\because }\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{-}\mathbf{b}\mathbf{}\mathbf{\pm }\mathbf{}\sqrt{{\mathbf{b}}^{\mathbf{2}}\mathbf{}\mathbf{-}\mathbf{}\mathbf{4}\mathbf{a}\mathbf{c}}}{\mathbf{2}\mathbf{a}}\right]$

$\Rightarrow$$\cos \theta =\frac{(\sqrt{3}+3\sqrt{3})}{4},\frac{(\sqrt{3}–3\sqrt{3})}{4}$

$\Rightarrow$$\cos \theta =\frac{\left(4\sqrt{3}\right)}{4},\frac{(-2\sqrt{3})}{4}$

$\Rightarrow$$\cos \theta =\sqrt{3},-\frac{\sqrt{3}}{2}$

As We know that $\cos \theta =\sqrt{3}$ is not possible.

$\because \cos \theta =-\frac{\sqrt{3}}{2}$

$\mathbf{\therefore }\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{5}\mathbf{\pi }}{\mathbf{6}}$

Hence, Option ‘C’ is Correct.