JEE Mathematics 2024 Question with Solution

Given: $4 \sin^2 x - 4 \cos^3 x + 9 - 4 \cos x = 0$ with $x \in [-2\pi, 2\pi]$.

Find: The number of solutions in the given interval.

Use the identity $\sin^2 x = 1 - \cos^2 x$ and substitute into the equation.

Let $y = \cos x$. Then $y \in [-1,1]$ and the equation becomes

Define

Check the values at the endpoints of the allowed interval:

Since the solution concludes that the expression remains positive on $[-1,1]$, there is no admissible value of $y = \cos x$ satisfying the equation. Hence there is no value of $x$ in $[-2\pi,2\pi]$ satisfying the given equation.

Therefore, the number of solutions is $0$. The correct option is D.

Given: $4\sin^2 x - 4\cos^3 x + 9 - 4\cos x = 0$.

Find: How many values of $x$ in $[-2\pi,2\pi]$ satisfy it.

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

Factoring out $-4$ gives

so we need

The solution states that after analyzing this equation for $x \in [-2\pi,2\pi]$, there are no real solutions satisfying it. Therefore the count of solutions is $0$, so the correct option is D.