Animated Solution for Mathematics - Functions: Let $f(x)=x^2$ and $g(x)=\sin x$ for all $x\in R$. Then the set of all $x$ satisfying $(f\circ g\circ g\circ f)(x)=(g\circ g\circ f)(x)$, where $(f\circ g)(x)=f(g(x))$, is

Imagine standing before a massive, tangled knot of ropes. If you try to pull every strand at once, you will only tighten the knot. But if you find the single loop that holds the structure together, the entire thing unravels with a gentle tug.

This problem is exactly like that knot. We are given the equation $(f\circ g\circ g\circ f)(x)=(g\circ g\circ f)(x)$, where $f(x)=x^2$ and $g(x)=\sin x$. At first glance, it looks like a chaotic nesting of functions, but let us breathe and look for the pattern.

Notice the repetition? The term $(g\circ g\circ f)(x)$ appears on both sides of the equation. Let us define a new function, $h(x)=(g\circ g\circ f)(x)$.

Suddenly, the entire equation transforms into something elegant:

Since we know $f(x)=x^2$, this becomes $(h(x){)}^2=h(x)$. This is not just an equation; it is a gateway. By rearranging it into $(h(x){)}^2-h(x)=0$, we factor it as $h(x)(h(x)-1)=0$.

We have successfully reduced a complex composition problem into two simple cases: $h(x)=0$ or $h(x)=1$.

Now, we must peel back the layers of our substitution. We defined $h(x)=g(g(f(x)))$. Substituting our known functions, we get:

Let us test our first case: $h(x)=1$. This implies $\sin (\sin (x^2))=1$.

For the sine of an angle to be $1$, the angle itself must be $\frac{\pi }{2}+2n\pi$. But wait—the inner angle is $\sin (x^2)$. The maximum value of $\sin (x^2)$ is $1$.

Since $\frac{\pi }{2}\approx 1.57$, it is impossible for $\sin (x^2)$ to ever reach $1.57$. The case $h(x)=1$ is a phantom; it exists in algebra but vanishes in the real world. We discard it with confidence.

Now for the second case: $h(x)=0$. This gives us $\sin (\sin (x^2))=0$.

For the sine of an angle to be $0$, the angle must be an integer multiple of $\pi$. Thus, $\sin (x^2)=n\pi$. Again, we must respect the boundaries of the sine function.

We know that $-1\le \sin (x^2)\le 1$. If $n$ is any integer other than $0$, the value of $n\pi$ will fall outside this range (e.g., $1\pi \approx 3.14$). Therefore, the only integer that works is $n=0$.

This leaves us with the beautiful, simple result: $\sin (x^2)=0$. This implies $x^2=k\pi$ for any non-negative integer $k$.

Taking the square root, we arrive at the final solution:

We have navigated the maze, ignored the traps, and arrived at the truth. Mathematics is not about brute force; it is about finding the path of least resistance. You have mastered the composition, and in doing so, you have mastered the problem.