Animated Solution for Mathematics - Trigonometry: The set of all $x$ in the interval $[0,\pi ]$ for which $2{\sin }^{2}x-3\sin x+1\ge 0$, is ..........

Welcome, warriors of JEE! Today, we are going to dismantle a trigonometric inequality that looks intimidating at first glance but is actually a beautiful exercise in pattern recognition.

We are tasked with finding all $x$ in the interval $[0,\pi ]$ such that:

At first, you might be tempted to overcomplicate things with trigonometric identities, but pause for a moment. Look at the structure; it is a quadratic in disguise!

By letting $t=\sin x$, the entire expression transforms into:

This is the power of substitution; it strips away the complexity and reveals the core algebraic challenge.

Now that we have our quadratic inequality $2t^2-3t+1\ge 0$, we need to find the values of $t$ that satisfy this. First, we factorize the quadratic.

Splitting the middle term, $-3t$, into $-2t$ and $-t$, we get:

Factoring by grouping, we arrive at:

This is where the wavy curve method becomes our most reliable ally. The critical points are $t=\frac{1}{2}$ and $t=1$.

Plotting these on a number line, we test the intervals. Since we need the expression to be greater than or equal to zero, our valid regions for $t$ are:

We cannot stop at $t$. We must return to our original variable, $x$. Substituting back, we have two conditions: $\sin x\le \frac{1}{2}$ and $\sin x\ge 1$.

Let's visualize this on the graph of $y=\sin x$ within the interval $[0,\pi ]$. For the first condition, $\sin x\le \frac{1}{2}$, we draw the horizontal line $y=\frac{1}{2}$.

The sine curve intersects this line at $x=\frac{\pi }{6}$ and $x=\frac{5\pi }{6}$. We want the parts of the curve that lie below or on this line, which gives us the intervals $[0,\frac{\pi }{6}]$ and $[\frac{5\pi }{6},\pi ]$.

Now, for the second condition, $\sin x\ge 1$. We know the maximum value of the sine function is $1$.

Thus, $\sin x\ge 1$ is only possible when $\sin x=1$. Within our interval $[0,\pi ]$, this occurs at exactly one point: $x=\frac{\pi }{2}$.

We have our pieces: the intervals $[0,\frac{\pi }{6}]$ and $[\frac{5\pi }{6},\pi ]$ from the first condition, and the discrete point $\frac{\pi }{2}$ from the second.

Combining these, we get the final solution set:

This problem is a perfect example of how JEE tests your ability to bridge the gap between algebraic manipulation and graphical intuition. Keep practicing, keep visualizing, and most importantly, keep falling in love with the elegance of mathematics!