Let S = {sin² 2θ : (sin⁴ θ + cos⁴ θ) x² + (sin 2θ)x + (sin⁶ θ + cos⁶ θ) = 0 has real roots}. If α an

Q: Wait, why do we define $t = \sin^2 2\theta$ instead of just $\sin 2\theta$?

Great observation! If we used $\sin 2\theta$, the discriminant would involve square roots or absolute values, making the inequality messy. By choosing $t = \sin^2 2\theta$, we map the trigonometric range $[-1, 1]$ to a clean, non-negative interval $[0, 1]$, which makes the quadratic inequality much easier to handle.

Q: Why can't we just use energy conservation or some other shortcut?

This isn't a physics problem, so energy conservation doesn't apply here! This is purely an algebraic constraint problem. The 'shortcut' is recognizing the symmetry in the coefficients $A$ and $C$—once you see that both can be written as functions of $\sin^2 2\theta$, the problem collapses into a simple quadratic inequality.

Q: How do I know if I should include the endpoints of the interval?

Always look at the inequality sign. We are given that the equation has 'real roots,' which implies $D \ge 0$. Because the inequality is non-strict, the boundary points where $D = 0$ are included in our set $S$. Always check if your calculated interval overlaps with the physical domain of the variable.

Q: Why did we reject the root $t_2 = 2 + \frac{2}{\sqrt{3}}$?

Excellent catch! Remember that $t = \sin^2 2\theta$. The maximum value of $\sin^2 2\theta$ is $1$. Since $t_2$ is greater than $1$, it is impossible for $\sin^2 2\theta$ to ever reach that value. We must always constrain our algebraic solutions by the physical reality of the trigonometric functions.

Q: Is there a trick to the final calculation $3((\alpha - 2)^2 + (\beta - 1)^2)$?

The trick is to not expand the squares immediately! Notice that $\beta = 1$ makes the second term vanish entirely. By keeping the expression in its factored form, you save time and reduce the chance of arithmetic errors.

Question

Let $S = \{\sin^2 2	heta : (\sin^4 	heta + \cos^4 	heta) x^2 + (\sin 2	heta)x + (\sin^6 	heta + \cos^6 	heta) = 0 	ext{ has real roots} \}$. If $\alpha$ and $\beta$ be the smallest and largest elements of the set $S$, respectively, then $3 ((\alpha - 2)^2 + (\beta - 1)^2)$ equals _________

Sigma Percentile · Accepted Answer

### The Symphony of Symmetry: Unlocking the Quadratic Constraint

Imagine you are standing before a complex, intimidating-looking quadratic equation: 
$$(\sin^4 	heta + \cos^4 	heta) x^2 + (\sin 2	heta)x + (\sin^6 	heta + \cos^6 	heta) = 0$$

At first glance, it looks like a mess of trigonometric powers. But in the world of JEE Advanced, complexity is often just a mask for hidden symmetry. Today, we are going to peel back that mask.

### Phase 1: The Discriminant as a Gateway

When we are told an equation has 'real roots,' our minds should immediately jump to the discriminant. For any quadratic $Ax^2 + Bx + C = 0$, the condition for real roots is $D = B^2 - 4AC \ge 0$.

This is our golden key. However, the coefficients $A$, $B$, and $C$ here are not constants; they are functions of $	heta$. If we try to solve this directly, we will drown in trigonometric identities. Instead, let us look for a common thread.

### Phase 2: The Power of Identities

Look closely at $A = \sin^4 	heta + \cos^4 	heta$ and $C = \sin^6 	heta + \cos^6 	heta$. These are classic symmetric expressions. We know that $\sin^2 	heta + \cos^2 	heta = 1$.

By squaring this, we get $\sin^4 	heta + \cos^4 	heta + 2\sin^2 	heta \cos^2 	heta = 1$, which simplifies to $A = 1 - 2\sin^2 	heta \cos^2 	heta$. Using the double angle identity $\sin 2	heta = 2\sin 	heta \cos 	heta$, we can rewrite this as:
$$A = 1 - \frac{1}{2}\sin^2 2	heta$$

Similarly, for $C$, we use the identity $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$ to find:
$$C = 1 - \frac{3}{4}\sin^2 2	heta$$

Suddenly, the entire problem is dancing to the tune of $t = \sin^2 2	heta$!

### Phase 3: The Transformation

By setting $t = \sin^2 2	heta$, we transform a trigonometric nightmare into a simple quadratic inequality. Our coefficients become $A = 1 - \frac{t}{2}$, $B^2 = t$, and $C = 1 - \frac{3t}{4}$.

The discriminant condition $D = B^2 - 4AC \ge 0$ now reads:
$$t - 4\left(1 - \frac{t}{2}ight)\left(1 - \frac{3t}{4}ight) \ge 0$$

Expand the product carefully:
$$4\left(1 - \frac{3t}{4} - \frac{t}{2} + \frac{3t^2}{8}ight) = 4 - 3t - 2t + \frac{3t^2}{2} = 4 - 5t + \frac{3t^2}{2}$$

Subtracting this from $t$, we get $t - (4 - 5t + \frac{3t^2}{2}) \ge 0$, which simplifies beautifully to:
$$3t^2 - 12t + 8 \le 0$$

### Phase 4: The Final Convergence

Solving $3t^2 - 12t + 8 = 0$ using the quadratic formula gives us $t = 2 \pm \frac{2}{\sqrt{3}}$. Since our parabola opens upward, the inequality holds between these two roots.

However, we must respect the physical domain of $t = \sin^2 2	heta$, which is $[0, 1]$. The intersection of $[2 - \frac{2}{\sqrt{3}}, 2 + \frac{2}{\sqrt{3}}]$ and $[0, 1]$ leaves us with the interval $[2 - \frac{2}{\sqrt{3}}, 1]$.

Thus, our smallest element is $\alpha = 2 - \frac{2}{\sqrt{3}}$ and our largest element is $\beta = 1$.

### The Grand Finale

Finally, we calculate $3((\alpha - 2)^2 + (\beta - 1)^2)$. Substituting our values, the second term vanishes because $\beta = 1$.

The first term becomes:
$$3\left(2 - \frac{2}{\sqrt{3}} - 2ight)^2 = 3\left(-\frac{2}{\sqrt{3}}ight)^2 = 3\left(\frac{4}{3}ight) = 4$$

The elegance of the final cancellation is the reward for your patience. You have mastered the symmetry, and the final answer is **4**.

Enter Numerical Value:

Visualized Solution

Setup and Discriminant Condition

Simplifying Coefficient $A$

Simplifying Coefficient $C$

Defining the Variable $t$

Setting up the Discriminant

Expanding and Simplifying

Solving the Quadratic Equation

Analyzing Roots and Domain

Identifying Alpha and Beta

Final Calculation

The Sigma Insight: Nature of Roots

The Symphony of Symmetry

Phase 1

Phase 2

Phase 3

Phase 4

The Grand Finale

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