Question

The smallest positive angle which satisfies the equation $2{\sin }^2\mathrm{\theta }+\sqrt{3}\cos \mathrm{\theta }+1=0$ is
(a) $\frac{5\pi }{6}$

(b) $\frac{2\pi }{3}$

(c) $\frac{\pi }{3}$

(d) $\frac{\pi }{6}$

Answer 1

(a) $\frac{5\pi }{6}$
Given:
$2{\sin }^2\theta +\sqrt{3}\cos \theta +1=0$
$$\begin{gathered} \Rightarrow 2(1-{\cos }^2\theta )+\sqrt{3}\cos \theta +1=0 \\ \Rightarrow 2-2{\cos }^2\theta +\sqrt{3}\cos \theta +1=0 \\ \Rightarrow 2{\cos }^2\theta -\sqrt{3}\cos \theta -3=0 \\ \Rightarrow 2{\cos }^2\theta -2\sqrt{3}\cos \theta +\sqrt{3}\cos \theta -3=0 \\ \Rightarrow 2\cos \theta (\cos \theta -\sqrt{3})+\sqrt{3}(\cos \theta -\sqrt{3})=0 \\ \Rightarrow (2\cos \theta +\sqrt{3})(\cos \theta -\sqrt{3})=0 \end{gathered}$$
$\Rightarrow 2\cos \theta +\sqrt{3}=0$ or, $\cos \theta -\sqrt{3}=0$

∴ $\cos \theta =-\frac{\sqrt{3}}{2}$ or, $\cos \theta =\sqrt{3}$ is not possible.
$$\begin{gathered} \Rightarrow \cos \theta =\cos \left(\frac{5\mathrm{\pi }}{6}\right) \\ \Rightarrow \theta =2n\mathrm{\pi }\pm \frac{5\mathrm{\pi }}{6},\mathit{}n\in Z \end{gathered}$$

For n = 0, the value of $\theta \mathrm{is}\pm \frac{5\mathrm{\pi }}{6}$.
Hence, the smallest positive angle is $\frac{5\mathrm{\pi }}{6}$.