Q41 / 75 · JEE Main 2026 (23 January Shift 1) · Mathematics
Number of solutions of $\sqrt{3} \cos 2 \theta+8 \cos \theta+3 \sqrt{3}=0, \theta \in[-3 \pi, 2 \pi]$ is:
- A0
- Correct answerB5
- C3
- D4
Solution
Using $\cos 2\theta = 2\cos^2\theta - 1$:
$2\sqrt{3}\cos^2\theta + 8\cos\theta + 2\sqrt{3} = 0 \Rightarrow \sqrt{3}\cos^2\theta + 4\cos\theta + \sqrt{3} = 0$.
$\cos\theta = \frac{-4 \pm 2}{2\sqrt{3}}$, giving $\cos\theta = -\frac{1}{\sqrt{3}}$ or $\cos\theta = -\sqrt{3}$ (rejected).
Let $\alpha = \cos^{-1}(-1/\sqrt{3}) \approx 2.186$ rad. General solution: $\theta = 2n\pi \pm \alpha$.
In $[-3\pi, 2\pi]$: $\theta = \alpha, -\alpha, 2\pi-\alpha, -2\pi+\alpha, -2\pi-\alpha$.
Total $= 5$ solutions.
$2\sqrt{3}\cos^2\theta + 8\cos\theta + 2\sqrt{3} = 0 \Rightarrow \sqrt{3}\cos^2\theta + 4\cos\theta + \sqrt{3} = 0$.
$\cos\theta = \frac{-4 \pm 2}{2\sqrt{3}}$, giving $\cos\theta = -\frac{1}{\sqrt{3}}$ or $\cos\theta = -\sqrt{3}$ (rejected).
Let $\alpha = \cos^{-1}(-1/\sqrt{3}) \approx 2.186$ rad. General solution: $\theta = 2n\pi \pm \alpha$.
In $[-3\pi, 2\pi]$: $\theta = \alpha, -\alpha, 2\pi-\alpha, -2\pi+\alpha, -2\pi-\alpha$.
Total $= 5$ solutions.
Hence, the correct answer is option: 'B'.