If $\theta \in[-2 \pi, 2 \pi]$, then the number of solutions of $2 \sqrt{2} \cos ^2 \theta+(2-\sqrt{6}) \cos \theta-\sqrt{3}=0$, is equal to:
Solution
<p>$$\begin{aligned}
& 2 \sqrt{2} \cos ^2 \theta+(2-\sqrt{6}) \cos \theta-\sqrt{3}=0 \\
& 2 \sqrt{2} \cos ^2 \theta+2 \cos \theta-\sqrt{6} \cos \theta-\sqrt{3}=0 \\
& (2 \cos \theta-\sqrt{3})(\sqrt{2} \cos \theta+1)=0 \\
& \Rightarrow \cos \theta=\frac{\sqrt{3}}{2} \text { or } \cos \theta=\frac{-1}{\sqrt{2}} \\
& \theta=\left\{\frac{-11 \pi}{6}, \frac{-5 \pi}{4}, \frac{-3 \pi}{4}, \frac{-\pi}{6}, \frac{\pi}{6}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{11 \pi}{6}\right\} \\
& \Rightarrow 8 \text { (solution) }
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Trigonometric Functions · Topic: Trigonometric Ratios and Identities
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