JEE Mathematics 2025 Question with Solution

Given: $\theta \in [-2\pi, 2\pi]$ and

$$2\sqrt{2} \cos^2\theta + (2 - \sqrt{6}) \cos\theta - \sqrt{3} = 0$$

Find: The total number of solutions in the given interval.

Let $x = \cos\theta$. Then the equation becomes

$$2\sqrt{2}x^2 + (2 - \sqrt{6})x - \sqrt{3} = 0$$

Using factorization shown in the solution,

$$(2x - \sqrt{3})(\sqrt{2}x + 1) = 0$$

So,

$$x = \frac{\sqrt{3}}{2} \quad \text{or} \quad x = -\frac{1}{\sqrt{2}}$$

Hence,

$$\cos\theta = \frac{\sqrt{3}}{2} \quad \text{or} \quad \cos\theta = -\frac{1}{\sqrt{2}}$$

For $\cos\theta = \frac{\sqrt{3}}{2}$, the solutions in $[-2\pi, 2\pi]$ are

$$\theta = \pm \frac{\pi}{6}, \ \pm \frac{11\pi}{6}$$

So this case gives $4$ solutions.

For $\cos\theta = -\frac{1}{\sqrt{2}}$, the solutions in $[-2\pi, 2\pi]$ are

$$\theta = \pm \frac{3\pi}{4}, \ \pm \frac{5\pi}{4}$$

So this case also gives $4$ solutions.

Therefore, the total number of solutions is

$$4 + 4 = 8$$

So, the correct option is C.

Given: The equation is quadratic in $\cos\theta$.

Find: The number of values of $\theta$ in $[-2\pi, 2\pi]$.

Instead of expanding the quadratic formula, observe that

$$2\sqrt{2}x^2 + (2 - \sqrt{6})x - \sqrt{3} = (2x - \sqrt{3})(\sqrt{2}x + 1)$$

with $x = \cos\theta$.

Thus,

$$\cos\theta = \frac{\sqrt{3}}{2} \quad \text{or} \quad \cos\theta = -\frac{1}{\sqrt{2}}$$

Each cosine value gives exactly $4$ solutions in the interval $[-2\pi, 2\pi]$.

Hence total number of solutions is

$$8$$

Therefore, the correct option is C.