JEE Mathematics 2025 Question with Solution

Given: $(4 - \sqrt{3}) \sin x - 2\sqrt{3} \cos^2 x = \frac{-4}{1 + \sqrt{3}}$ and $x \in \left[-2\pi, \frac{5\pi}{2}\right]$.

Find: The number of solutions in the given interval.

First rationalize the right-hand side:

Now substitute $\cos^2 x = 1 - \sin^2 x$:

Let $y = \sin x$. Then:

Using the quadratic formula,

Since $19 + 8\sqrt{3} = (4 + \sqrt{3})^2$,

So the possible values are:

But $\frac{-2}{\sqrt{3}} < -1$, so it is not possible for sine. Hence we only need to solve $\sin x = \frac{1}{2}$.

The general solutions are:

Checking values in $\left[-2\pi, \frac{5\pi}{2}\right]$:

• For $n = -2$, $x = -\frac{11\pi}{6}$
• For $n = -1$, $x = -\frac{7\pi}{6}$
• For $n = 0$, $x = \frac{\pi}{6}$
• For $n = 1$, $x = \frac{5\pi}{6}$
• For $n = 2$, $x = \frac{13\pi}{6}$
• For $n = 3$, $x = \frac{17\pi}{6} > \frac{5\pi}{2}$, so it is excluded.

Thus the solutions in the interval are $-\frac{11\pi}{6}, -\frac{7\pi}{6}, \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}$. Therefore, the number of solutions is $5$ and the correct option is D.

Given: $(4 - \sqrt{3}) \sin x - 2\sqrt{3} \cos^2 x = \frac{-4}{1 + \sqrt{3}}$.

Find: How many values of $x$ in $\left[-2\pi, \frac{5\pi}{2}\right]$ satisfy the equation.

The key idea is to convert the equation into a polynomial in one trigonometric function. After rationalizing,

so the equation becomes

Using $\cos^2 x = 1 - \sin^2 x$,

which simplifies to

This is a quadratic in $\sin x$.

Solving it gives $\sin x = \frac{1}{2}$ or $\sin x = -\frac{2}{\sqrt{3}}$. Since sine cannot lie outside $[-1,1]$, only $\sin x = \frac{1}{2}$ is valid.

Now count all points in the interval where sine equals $\frac{1}{2}$. In each full period, the solutions are

Shifting these by multiples of $2\pi$ and keeping only those inside $\left[-2\pi, \frac{5\pi}{2}\right]$ gives exactly five values.

Therefore, the required number of solutions is $5$. Hence, the correct option is D.