Animated Solution for Mathematics - Trigonometry: Find all the solution of $4{\cos }^{2}x\sin x-2{\sin }^{2}x=3\sin x$.

Welcome, my dear student. Today, we are not just solving an equation; we are performing a delicate dance with trigonometry. When you look at an equation like $4{\cos }^{2}x\sin x-2{\sin }^{2}x=3\sin x$, it is easy to feel overwhelmed.

It looks messy, doesn't it? We have a mix of ${\cos }^{2}x$, ${\sin }^{2}x$, and $\sin x$. But in the world of JEE Advanced, complexity is just a mask for elegance waiting to be revealed.

The Golden Rule of Factoring

Let us begin by addressing the most common mistake students make. You see $\sin x$ everywhere, and your hand immediately reaches to divide the entire equation by $\sin x$. Stop!

If you divide by $\sin x$, you are effectively killing the solutions where $\sin x=0$. You are throwing away valid answers before you have even begun.

Instead, let us bring everything to one side. We transform the equation into:

Now, we factor out $\sin x$. This is the moment of clarity. We are left with:

This is beautiful. We have split our problem into two distinct paths. Either $\sin x=0$, or the expression inside the bracket is zero. We have already secured our first set of solutions: $x=n\pi$.

The Transformation

Now, look at the bracket: $4{\cos }^{2}x-2\sin x-3=0$. We have a mixture of cosine and sine. We cannot solve this as it stands.

We need homogeneity. We need everything in the language of sine. We invoke the most powerful tool in our trigonometric arsenal: the Pythagorean identity, ${\cos }^{2}x=1-{\sin }^{2}x$.

Substituting this into our equation, we get:

Let us expand this carefully. Do not rush. A single sign error here will haunt you for the rest of the problem. We get:

Simplifying the constants, we arrive at:

To make our lives easier, let us multiply by $-1$ to make the leading coefficient positive. We now have a clean, standard quadratic equation:

The Quadratic Engine

We have already handled $\sin x=0$. Now, we must conquer the quadratic factor: $4{\sin }^{2}x+2\sin x-1=0$. This does not factorize easily by inspection.

When the path is not obvious, we rely on the quadratic formula:

Substituting $a=4$, $b=2$, and $c=-1$, we find:

Simplifying $\sqrt{20}$ to $2\sqrt{5}$, we get:

The Hidden Gems

Here is where the JEE Advanced examiner tests your intuition. You have two values for $\sin x$: $\frac{\sqrt{5}-1}{4}$ and $\frac{-\sqrt{5}-1}{4}$.

Do these numbers look familiar? They should. The value $\frac{\sqrt{5}-1}{4}$ is the exact value of $\sin 18^{\circ }$, or $\sin \frac{\pi }{10}$. The value $\frac{-\sqrt{5}-1}{4}$ is the value of $\sin (-54^{\circ })$, or $\sin \left(-\frac{3\pi }{10}\right)$.

Recognizing these special angles is the difference between a good student and a topper. It allows us to write the general solutions using the standard form $x=n\pi +(-1{)}^n\alpha$.

For $\alpha =\frac{\pi }{10}$, we have $x=n\pi +(-1{)}^n\frac{\pi }{10}$. For $\alpha =-\frac{3\pi }{10}$, we have $x=n\pi +(-1{)}^n\left(-\frac{3\pi }{10}\right)$.

We started with a daunting equation, and through systematic factoring, identity substitution, and recognizing special values, we have dismantled it completely.

The solutions are $x=n\pi$, $x=n\pi +(-1{)}^n\frac{\pi }{10}$, and $x=n\pi +(-1{)}^n\left(-\frac{3\pi }{10}\right)$.

Remember, my student: math is not about memorizing steps. It is about recognizing the structure. When you see mixed trigonometric functions, think of identities. When you see a quadratic, think of the formula. And always, always look for the common factor first. You have the tools; now go forth and conquer.