Animated Solution for Mathematics - Trigonometry: For $x\in (0,\pi )$, the equation $\sin x+2\sin 2x-\sin 3x=3$ has

Welcome, future engineers. Today, we are not just solving a trigonometric equation; we are embarking on a detective mission. We are looking at the equation $\sin x+2\sin 2x-\sin 3x=3$ for $x\in (0,\pi )$.

At first glance, it looks like a standard problem. You might be tempted to start rearranging terms, squaring both sides, or perhaps trying to isolate $x$. But stop. Take a breath.

In the JEE Advanced arena, the most powerful tool in your arsenal is not just calculation—it is intuition. Before we dive into the algebra, we must ask: Is it even possible for this expression to reach 3?

To understand the behavior of this function, we need to speak the same language. The equation is cluttered with multiple angles: $2x$ and $3x$. These are the obstacles. We need to break them down into the fundamental unit: $x$.

Recall your double angle identity: $\sin 2x=2\sin x\cos x$. Then, recall the triple angle identity: $\sin 3x=3\sin x-4{\sin }^{3}x$. These are the keys to the kingdom.

By substituting these into our original equation, we transform a complex, multi-angle expression into a single-angle polynomial:

Look at that! It looks a bit messy, but it is entirely in terms of $\sin x$ and $\cos x$. Let's simplify. We have $\sin x-3\sin x$, which gives us $-2\sin x$.

The term $4\sin x\cos x$ remains, and the negative sign distributes to the $-4{\sin }^{3}x$, turning it into $+4{\sin }^{3}x$. Our equation now reads:

Now, observe the left-hand side. Every single term contains a $\sin x$. This is the moment where the problem begins to yield. Let's factor out $\sin x$:

We are getting closer. But we have a mix of ${\sin }^{2}x$ and $\cos x$ inside the bracket. This is a recipe for confusion. We need uniformity.

Using the Pythagorean identity, ${\sin }^{2}x=1-{\cos }^{2}x$, we can rewrite the bracket entirely in terms of $\cos x$:

Simplifying the expression inside the bracket gives us:

This is where the "JEE Advanced" mindset kicks in. Most students would try to solve for $x$ here. Instead, let's define a function $f(x)=\sin x(2+4\cos x-4{\cos }^{2}x)$. We want to know if $f(x)$ can ever equal 3.

Let $t=\cos x$. Since $x\in (0,\pi )$, $t$ ranges from $-1$ to $1$. The quadratic part is $g(t)=2+4t-4t^2$.

Let's find the maximum of this quadratic. Differentiating $g(t)$ with respect to $t$, we get $g^{\prime }(t)=4-8t$. Setting this to zero, we find the critical point at $t=\frac{1}{2}$.

Plugging $t=\frac{1}{2}$ back into $g(t)$, we get:

So, the quadratic part has a maximum value of 3. But wait! The full expression is $\sin x\cdot g(t)$. For the whole expression to be 3, we would need $\sin x$ to be 1 at the exact same moment that $g(t)$ is 3.

Let's check the conditions. $g(t)=3$ happens when $t=\cos x=\frac{1}{2}$. If $\cos x=\frac{1}{2}$, then $x=\frac{\pi }{3}$. At $x=\frac{\pi }{3}$, what is $\sin x$? It is $\frac{\sqrt{3}}{2}$, which is approximately $0.866$.

So, at the point where the quadratic part is maximized, the value of our function is:

$2.598$ is strictly less than 3. Even if we try to maximize $\sin x$ by setting $x=\frac{\pi }{2}$ (where $\sin x=1$), the quadratic part becomes $g(0)=2$, and the product is $1\times 2=2$, which is also less than 3.

We have tested the peaks. We have analyzed the boundaries. The function $f(x)$ simply never reaches the height of 3.

It is a beautiful, continuous curve that dances below our target line, never touching it. Therefore, we can confidently conclude that there is no solution.