Animated Solution for Mathematics - Trigonometry: The sum of all values of $\theta \in [0,2\pi ]$ satisfying $2{\sin }^2\theta =\cos 2\theta$ and $2{\cos }^2\theta =3\sin \theta$ is

Welcome, fellow traveler on the path to JEE mastery! Today, we are not just solving a trigonometry problem; we are orchestrating a symphony of identities.

We have two equations, both vying for our attention, and our goal is to find the values of $\theta$ that satisfy both simultaneously within the interval $[0,2\pi ]$.

Imagine you are trying to translate two different languages into one. Our first equation is $2{\sin }^2\theta =\cos 2\theta$, and our second is $2{\cos }^2\theta =3\sin \theta$.

They are currently speaking different dialects—some terms are in sine, some in cosine, and some are double angles. To make them communicate, we must unify them.

For the first equation, we invoke the double angle identity: $\cos 2\theta =1-2{\sin }^2\theta$. Substituting this into our first equation gives us:

With a quick rearrangement, we get $4{\sin }^2\theta =1$, which simplifies beautifully to:

This tells us that $\sin \theta$ must be either $\frac{1}{2}$ or $-\frac{1}{2}$.

Now, let us turn our attention to the second equation: $2{\cos }^2\theta =3\sin \theta$. We use the Pythagorean identity, ${\cos }^2\theta =1-{\sin }^2\theta$, to transform this into:

Expanding this, we arrive at $2-2{\sin }^2\theta =3\sin \theta$. If we move everything to one side, we get the quadratic equation:

Don't let the trigonometry scare you! If you treat $\sin \theta$ as a variable, say $x$, you are simply solving $2x^2+3x-2=0$.

Factorizing this by splitting the middle term, we get $(2\sin \theta -1)(\sin \theta +2)=0$. This yields two potential roots: $\sin \theta =\frac{1}{2}$ or $\sin \theta =-2$.

Here is where the JEE tests your vigilance. We have two sets of potential solutions.

From the first equation, $\sin \theta \in \{\frac{1}{2},-\frac{1}{2}\}$. From the second, $\sin \theta \in \{\frac{1}{2},-2\}$.

But wait! We must reject $\sin \theta =-2$ because the sine function is bounded between $-1$ and $1$. Thus, the only valid solution that satisfies both equations is $\sin \theta =\frac{1}{2}$.

Now, we look at the unit circle. Where is $\sin \theta =\frac{1}{2}$?

In the first quadrant, we have $\theta =\frac{\pi }{6}$. In the second quadrant, by symmetry, we have $\theta =\pi -\frac{\pi }{6}=\frac{5\pi }{6}$.

These are our two valid angles. The problem asks for the sum of all such values:

And there you have it! Through careful identity substitution, quadratic factorization, and a keen eye for the range of trigonometric functions, we have arrived at the final answer of $\pi$.