JEE Mains · Maths · STD 11 - Trigonometrical equations
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:
- A 12
- B 6
- C 8
- D 10
Answer & Solution
Correct Answer
(C) 8
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & 2 \sqrt{2} \cos ^2 \theta+2 \cos \theta-\sqrt{6} \cos \theta-\sqrt{3}=0 \\ & (2 \cos \theta-\sqrt{3})(\sqrt{2} \cos \theta+1)=0 \\ & \cos \theta=\frac{\sqrt{3}}{2}, \frac{-1}{\sqrt{2}}\end{aligned}\) Number of solution \(=8\)
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