Trigonometric Equations – General Solution – JEE Main 23 Jan 2026 Shift 1

Solution:

We use the double angle identity $\cos 2\theta = 2\cos^2\theta – 1$.
The equation becomes:
$$ \sqrt{3}(2\cos^2\theta – 1) + 8\cos\theta + 3\sqrt{3} = 0 $$

$$ 2\sqrt{3}\cos^2\theta – \sqrt{3} + 8\cos\theta + 3\sqrt{3} = 0 $$

$$ 2\sqrt{3}\cos^2\theta + 8\cos\theta + 2\sqrt{3} = 0 $$

Divide by 2:
$$ \sqrt{3}\cos^2\theta + 4\cos\theta + \sqrt{3} = 0 $$

This is a quadratic in $\cos\theta$. Factorize it:
$$ \sqrt{3}\cos^2\theta + 3\cos\theta + \cos\theta + \sqrt{3} = 0 $$

$$ \sqrt{3}\cos\theta(\cos\theta + \sqrt{3}) + 1(\cos\theta + \sqrt{3}) = 0 $$

$$ (\sqrt{3}\cos\theta + 1)(\cos\theta + \sqrt{3}) = 0 $$

This gives two cases:
1) $\cos\theta = -\sqrt{3}$ (No solution, since $|\cos\theta| \le 1$)
2) $\cos\theta = -\frac{1}{\sqrt{3}}$

We need to count the solutions for $\cos\theta = -\frac{1}{\sqrt{3}}$ in the interval $[-3\pi, 2\pi]$.
Since $-\frac{1}{\sqrt{3}} \approx -0.577$, the line $y = -0.577$ intersects the cosine graph twice in every period of $2\pi$.

Interval Breakdown:
– In $[0, 2\pi]$: 2 solutions (2nd and 3rd quadrants).
– In $[-2\pi, 0]$: 2 solutions.
– In $[-3\pi, -2\pi]$: This covers the range from $-3\pi$ to $-2\pi$ (equivalent to $\pi$ to $2\pi$). The cosine curve goes from $-1$ to $1$. The value $-\frac{1}{\sqrt{3}}$ is crossed once in this half-period (specifically in the 3rd quadrant equivalent).

Total solutions = $2 + 2 + 1 = 5$.

Ans. (2)