Animated Solution for Mathematics - Trigonometry: The number of values of $x$ in the interval $[0,5\pi ]$ satisfying the equation $3{\sin }^{2}x-7\sin x+2=0$ is

Welcome, future engineers! Today, we are going to demystify a problem that often trips up students in the JEE Advanced exam. We are looking at the equation $3{\sin }^{2}x-7\sin x+2=0$ over the interval $[0,5\pi ]$.

At first glance, it looks like a trigonometric nightmare. But let's peel back the layers. The first thing you must notice is the structure; it is a quadratic equation in disguise.

By substituting $t=\sin x$, the equation transforms into:

This is the beauty of mathematics—we take a complex problem and map it onto a familiar structure. Factorizing this, we get $(3t-1)(t-2)=0$. This leads us to $t=\frac{1}{3}$ or $t=2$.

Respecting the Bounds

Now, here is where the conceptual trap lies. Many students rush to solve for $x$ in both cases. But remember the fundamental property of the sine function: $-1\le \sin x\le 1$.

The value $t=2$ is impossible! It falls outside the range of the sine function. So, we discard it.

We are left with the elegant task of solving $\sin x=\frac{1}{3}$. This is a crucial moment in your problem-solving journey. Never blindly solve an equation; always check if the solutions are physically or mathematically valid within the domain of the function.

The Power of the Graph

Instead of calculating exact values, we use the power of visualization. We are working in the interval $[0,5\pi ]$. Let's break this down.

Imagine the sine wave oscillating on your graph paper. We are looking for the intersection of this wave with the horizontal line $y=\frac{1}{3}$.

In the first cycle $[0,2\pi ]$, the sine wave is positive in the first and second quadrants. The line $y=\frac{1}{3}$ cuts the wave twice.

In the second cycle $[2\pi ,4\pi ]$, the pattern repeats, giving us two more solutions. Finally, in the interval $[4\pi ,5\pi ]$, we are looking at the first half of the third cycle (the interval $[4\pi ,5\pi ]$ corresponds to the first and second quadrants of the unit circle).

Since $\sin x=\frac{1}{3}$ is positive, it intersects the sine wave twice in the interval $[4\pi ,5\pi ]$. Adding these up, we get $2+2+2=6$ solutions.

Final Answer: There are 6 solutions in the given interval. Always remember: visualize, check your constraints, and break complex intervals into manageable cycles. You have got this!