Animated Solution for Mathematics - Trigonometry: The sum of all values of $\theta \in (0,\frac{\pi }{2})$ satisfying ${\sin }^{2}2\theta +{\cos }^{4}2\theta =\frac{3}{4}$ is :

Welcome, fellow traveler on the JEE journey! Today, we are going to unravel a problem that might look like a tangled mess of powers and ratios, but beneath the surface, it is a beautifully choreographed dance of trigonometric identities.

We are tasked with finding the sum of all values of $\theta$ in the interval $(0,\frac{\pi }{2})$ that satisfy the equation:

The Transformation

Imagine you are standing in front of this equation. It feels a bit disjointed, doesn't it? We have a ${\sin }^{2}2\theta$ and a ${\cos }^{4}2\theta$.

In the world of trigonometry, we crave uniformity. We want to speak one language. Since we have a ${\cos }^{4}2\theta$ term, it is far more elegant to convert the sine term into cosine.

We reach into our toolkit and pull out the most fundamental identity: ${\sin }^{2}A=1-{\cos }^{2}A$. By setting $A=2\theta$, we can rewrite our equation as:

Suddenly, the chaos subsides. We are now speaking entirely in the language of cosine.

The Algebraic Dance

Now, let's clean up the workspace. We have fractions, and nobody likes fractions in the middle of a derivation. Let's multiply the entire equation by $4$ to clear the denominator:

Expanding this gives us $4-4{\cos }^{2}2\theta +4{\cos }^{4}2\theta =3$. If we bring the $3$ over to the left, we get:

Take a deep breath and look at this. It is a quadratic equation in disguise! If you let $x={\cos }^{2}2\theta$, the equation becomes $4x^2-4x+1=0$.

Does that look familiar? It is a perfect square! It is of the form $(2x-1{)}^2=0$. So, our equation collapses into:

This is the moment where the math rewards your patience.

The Geometric Reality

Since the square is zero, the base must be zero: $2{\cos }^{2}2\theta -1=0$, which simplifies to ${\cos }^{2}2\theta =\frac{1}{2}$.

Now, here is the trap that catches many students: taking the square root. When you take the square root of ${\cos }^{2}2\theta =\frac{1}{2}$, you get:

Do not discard the negative root! We must now look at our domain. We were given $\theta \in (0,\frac{\pi }{2})$, which means $2\theta \in (0,\pi )$.

On the unit circle, this corresponds to the entire upper half-plane. We are looking for angles where the x-coordinate is $\frac{1}{\sqrt{2}}$ or $-\frac{1}{\sqrt{2}}$.

In the first quadrant, $\cos 2\theta =\frac{1}{\sqrt{2}}$ gives $2\theta =\frac{\pi }{4}$. In the second quadrant, $\cos 2\theta =-\frac{1}{\sqrt{2}}$ gives $2\theta =\frac{3\pi }{4}$. Both are valid!

We are almost there. We have $2\theta =\frac{\pi }{4}$ and $2\theta =\frac{3\pi }{4}$. Dividing by $2$, we find our values for $\theta$:

The question asks for the sum of these values. Adding them together:

And there it is. A beautiful, clean result. You navigated the identities, solved the quadratic, respected the domain, and arrived at the truth. The final sum is $\frac{\pi }{2}$.