Question

The smallest positive angle which satisfies the equation ​

\[2 \sin^2 x + \sqrt{3} \cos x + 1 = 0\] is

Options

• \[\frac{5\pi}{6}\]

   

• \[\frac{2\pi}{3}\]

   

• \[\frac{\pi}{3}\]

   

• \[\frac{\pi}{6}\]

   

Answer 1

\[\frac{5\pi}{6}\]
Given:
\[2 \sin^2 x + \sqrt{3}\cos x + 1 = 0\]
\[\Rightarrow 2 (1 - \cos^2 x) + \sqrt{3} \cos x + 1 = 0\]
\[ \Rightarrow 2 - 2 \cos^2 x + \sqrt{3} \cos x + 1 = 0\]
\[ \Rightarrow 2 \cos^2 x - \sqrt{3} \cos x - 3 = 0\]
\[ \Rightarrow 2 \cos^2 x - 2\sqrt{3} \cos x + \sqrt{3} \cos x - 3 = 0\]
\[ \Rightarrow 2 \cos x (\cos x - \sqrt{3}) + \sqrt{3} (\cos x - \sqrt{3}) = 0\]
\[ \Rightarrow (2 \cos x + \sqrt{3}) (\cos x - \sqrt{3}) = 0\]

\[\Rightarrow 2 \cos x + \sqrt{3} = 0\] or,

\[\cos x - \sqrt{3} = 0\]

∴ \[\cos x = - \frac{\sqrt{3}}{2}\] or,

\[\cos x = \sqrt{3}\] is not possible.

\[\Rightarrow \cos x = \cos\left( \frac{5\pi}{6} \right)\]
\[ \Rightarrow x = 2n\pi \pm \frac{5\pi}{6} , n \in Z\]

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