Animated Solution for Mathematics - Quadratic Equations: Find the range of values of $t$ for which $2\sin t=\frac{1-2x+5x^2}{3x^2-2x-1}$, $t\in [-\pi /2,\pi /2]$.

Imagine you are standing at the crossroads of algebra and trigonometry. You have been given a problem that looks like a simple equation:

We are looking for the range of $t$ in the interval $[-\frac{\pi }{2},\frac{\pi }{2}]$ such that $x$ remains a real number. This is the beauty of JEE Advanced problems—they force you to see the hidden dependencies.

Let us simplify our life. Let $y=2\sin t$. Now, our problem transforms into finding the range of $y$ such that the equation $y=\frac{5x^2-2x+1}{3x^2-2x-1}$ yields real values for $x$.

To solve this, we cross-multiply:

Expanding this, we get $3y{x}^2-2yx-y=5x^2-2x+1$. Now, gather all terms on one side to form a quadratic in $x$:

For $x$ to be real, the discriminant $D$ must be greater than or equal to zero. Using $D=B^2-4AC$, we substitute our coefficients:

When we expand this, we get $4(y-1{)}^2+4(3y-5)(y+1)\ge 0$. Simplifying this, we arrive at:

This is a beautiful, clean inequality: $y^2-y-1\ge 0$. The roots of $y^2-y-1=0$ are $\frac{1\pm \sqrt{5}}{2}$.

Since the inequality is $\ge 0$, $y$ must lie outside these roots. This creates a 'forbidden zone' between $\frac{1-\sqrt{5}}{2}$ and $\frac{1+\sqrt{5}}{2}$. Our function simply cannot exist there.

Now, we bring trigonometry back into the fold. We know $y=2\sin t$. Substituting this back, we get:

Dividing by 2, we find:

Here is where the JEE expertise shines. You must recognize these values. Since $\frac{\sqrt{5}-1}{4}=\sin (18^{\circ })$, it follows that $\frac{1-\sqrt{5}}{4}=\sin (-18^{\circ })=\sin (-\frac{\pi }{10})$. Similarly, $\frac{1+\sqrt{5}}{4}=\sin (54^{\circ })=\sin (\frac{3\pi }{10})$.

We are working in the interval $t\in [-\frac{\pi }{2},\frac{\pi }{2}]$. In this range, $\sin t$ is strictly increasing, making our mapping straightforward.

The condition $\sin t\le \sin (-\frac{\pi }{10})$ implies $t\in [-\frac{\pi }{2},-\frac{\pi }{10}]$. The condition $\sin t\ge \sin (\frac{3\pi }{10})$ implies $t\in [\frac{3\pi }{10},\frac{\pi }{2}]$.

Combining these, we arrive at our final, elegant solution:

You have successfully navigated the intersection of algebra and trigonometry. Take a moment to appreciate the symmetry of the result.