JEE Mathematics 2026 Question with Solution

Given: $\sqrt{3} \cos 2\theta + 8 \cos \theta + 3\sqrt{3} = 0$ with $\theta \in [-3\pi, 2\pi]$.

Find: The number of solutions in the given interval.

Use the identity

$$\cos 2\theta = 2\cos^2 \theta - 1$$

Substituting into the equation,

$$\sqrt{3}(2\cos^2 \theta - 1) + 8\cos \theta + 3\sqrt{3} = 0$$ $$2\sqrt{3}\cos^2 \theta + 8\cos \theta + 2\sqrt{3} = 0$$

Dividing by $2$,

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

Now solve this quadratic in $\cos \theta$:

$$\cos \theta = \frac{-4 \pm \sqrt{16 - 12}}{2\sqrt{3}} = \frac{-4 \pm 2}{2\sqrt{3}}$$

So the roots are

$$\cos \theta = -\frac{1}{\sqrt{3}} \quad \text{and} \quad \cos \theta = -\sqrt{3}$$

Since $|-\sqrt{3}| > 1$, this value is not possible for cosine, so it is rejected.

Thus we only need to count solutions of

$$\cos \theta = -\frac{1}{\sqrt{3}}$$

In each complete interval of length $2\pi$, this has exactly $2$ solutions because the cosine value lies strictly between $-1$ and $1$.

Now count in $[-3\pi, 2\pi]$:

1. In $[-3\pi, -\pi]$: $2$ solutions.
2. In $[-\pi, \pi]$: $2$ solutions.
3. In $[\pi, 2\pi]$: $1$ solution.

Therefore, the total number of solutions is $5$.

The correct option is D.

Given: $\cos \theta = -\frac{1}{\sqrt{3}}$ after simplification.

Find: How many values of $\theta$ lie in $[-3\pi, 2\pi]$.

Because $\left|-\frac{1}{\sqrt{3}}\right| < 1$, the equation $\cos \theta = k$ gives exactly $2$ solutions in every full period of length $2\pi$.

The interval $[-3\pi, 2\pi]$ consists of two full periods,

$$[-3\pi, -\pi] \quad \text{and} \quad [-\pi, \pi]$$

contributing $2 + 2 = 4$ solutions, and one extra half-period,

$$[\pi, 2\pi]$$

which contributes $1$ more solution.

Hence the total number of solutions is $5$, so the correct option is D.