Animated Solution for Mathematics - Trigonometry: The number of solutions of the equation $1+{\sin }^{4}x={\cos }^{2}3x$, $x\in [-\frac{5\pi }{2},\frac{5\pi }{2}]$ is :

Solving the Impossible

Welcome, future engineers. Today, we are going to dismantle a problem that, at first glance, looks like a nightmare of trigonometric expansion. You see the equation $1+{\sin }^{4}x={\cos }^{2}3x$ and your brain immediately starts searching for identities.

You might think, 'Should I expand ${\sin }^{4}x$? Should I use the triple angle formula for $\cos 3x$?' Stop. Take a breath. In the JEE Advanced arena, the most complex-looking problems often have the most elegant, simple solutions. This is not a problem of expansion; it is a problem of observation.

The Range Analysis

Imagine you are standing on a mountain looking at two different paths. One path is the Left-Hand Side (LHS), defined by $f(x)=1+{\sin }^{4}x$. The other is the Right-Hand Side (RHS), defined by $g(x)={\cos }^{2}3x$.

We want to find where these two paths cross. Instead of calculating every point, let's look at their boundaries.

Consider the LHS: $1+{\sin }^{4}x$. We know that for any real number $x$, the sine function $\sin x$ oscillates between $-1$ and $1$. Therefore, ${\sin }^{2}x$ is always between $0$ and $1$.

If we square that again to get ${\sin }^{4}x$, the range remains $[0,1]$. Now, add $1$ to this expression. The minimum value of ${\sin }^{4}x$ is $0$, so the minimum value of our LHS is $1+0=1$.

The maximum value of ${\sin }^{4}x$ is $1$, so the maximum value of our LHS is $1+1=2$. Thus, we have established that $LHS\ge 1$.

Now, look at the RHS: ${\cos }^{2}3x$. The cosine function, regardless of the argument $3x$, is bounded by $[-1,1]$. When you square it, the range becomes $[0,1]$.

This means the RHS can never exceed $1$. So, we have $RHS\le 1$.

The Boundary Condition

This is the 'Aha!' moment. We have a mathematical standoff. We have an equation where the LHS is always greater than or equal to $1$, and the RHS is always less than or equal to $1$.

If you have two numbers, $A$ and $B$, where $A\ge 1$ and $B\le 1$, and you are told that $A=B$, there is only one logical conclusion: both $A$ and $B$ must be exactly $1$. There is no other way for them to meet. They are forced to intersect at the boundary value of $1$.

This transforms our difficult trigonometric equation into a system of two much simpler equations:

Solving the System

Let's solve the first one: $1+{\sin }^{4}x=1$. Subtracting $1$ from both sides gives us ${\sin }^{4}x=0$, which implies $\sin x=0$.

We know from the unit circle that $\sin x=0$ at integer multiples of $\pi$. So, the general solution is $x=n\pi$, where $n\in \mathbb{Z}$.

Now, we must verify if these solutions satisfy the second condition: ${\cos }^{2}3x=1$. Let's substitute $x=n\pi$ into the RHS. We get ${\cos }^2(3n\pi )$.

Since $n$ is an integer, $3n$ is also an integer. The cosine of any integer multiple of $\pi$ is either $1$ or $-1$. When we square that result, we get $1^2=1$ or $(-1{)}^2=1$. It works perfectly! Every solution of the first equation is a valid solution for the second.

Counting the Solutions

Finally, we need to find how many of these solutions fall within the given interval $x\in [-\frac{5\pi }{2},\frac{5\pi }{2}]$. Let's convert the interval to decimals to make it easier to visualize: $[-\frac{5\pi }{2},\frac{5\pi }{2}]$ is approximately $[-2.5\pi ,2.5\pi ]$.

We test integer values for $n$:

- If $n=0$, $x=0$. (Inside) - If $n=1$, $x=\pi$. (Inside) - If $n=-1$, $x=-\pi$. (Inside) - If $n=2$, $x=2\pi$. (Inside) - If $n=-2$, $x=-2\pi$. (Inside) - If $n=3$, $x=3\pi$. (Outside, as $3>2.5$) - If $n=-3$, $x=-3\pi$. (Outside, as $-3<-2.5$)

Counting them up, we have $\{-2\pi ,-\pi ,0,\pi ,2\pi \}$. That gives us exactly 5 solutions.

See how we avoided the trap? By using range analysis, we bypassed the need for complex identities and algebraic manipulation.

This is the hallmark of a JEE Advanced topper: the ability to step back, analyze the structure of the problem, and choose the most efficient path. Keep this technique in your toolkit—whenever you see bounded functions in an equation, always check their ranges first. You might just find the answer staring back at you.