Question

The smallest positive angle which satisfies the equation ​$2{\sin }^{2}x+\sqrt{3}\cos x+1=0$ 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:
$2{\sin }^{2}x+\sqrt{3}\cos x+1=0$
$$\begin{gathered} \Rightarrow 2(1-{\cos }^{2}x)+\sqrt{3}\cos x+1=0 \\ \Rightarrow 2-2{\cos }^{2}x+\sqrt{3}\cos x+1=0 \\ \Rightarrow 2{\cos }^{2}x-\sqrt{3}\cos x-3=0 \\ \Rightarrow 2{\cos }^{2}x-2\sqrt{3}\cos x+\sqrt{3}\cos x-3=0 \\ \Rightarrow 2\cos x(\cos x-\sqrt{3})+\sqrt{3}(\cos x-\sqrt{3})=0 \\ \Rightarrow (2\cos x+\sqrt{3})(\cos x-\sqrt{3})=0 \end{gathered}$$
$\Rightarrow 2\cos x+\sqrt{3}=0$ or, $\cos x-\sqrt{3}=0$

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

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