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જો $\theta \in[-2 \pi, 2 \pi]$, હોય તો $2 \sqrt{2} \cos ^2 \theta+(2-\sqrt{6}) \cos \theta-\sqrt{3}=0$ ના ઉકેલની સંખ્યા મેળવો.
A$12$
B$6$
C$8$
D$10$
(JEE MAIN-2025)
Solution
$2 \sqrt{2} \cos ^2 \theta+2 \cos \theta-\sqrt{6} \cos \theta-\sqrt{3}=0$
$(2 \cos \theta-\sqrt{3})(\sqrt{2} \cos \theta+1)=0$
$\cos \theta=\frac{\sqrt{3}}{2}, \frac{-1}{\sqrt{2}}$
Number of solution $=8$
$(2 \cos \theta-\sqrt{3})(\sqrt{2} \cos \theta+1)=0$
$\cos \theta=\frac{\sqrt{3}}{2}, \frac{-1}{\sqrt{2}}$
Number of solution $=8$
Standard 11
Mathematics
