Given functions $f(x)=x^2, \qquad g(x)=\sin x$

We need to solve $(f\circ g\circ g\circ f)(x)=(g\circ g\circ f)(x).$

Compute each composition carefully

Start from inside:

• $f(x)=x^2$
• $g(f(x))=\sin(x^2)$
• $g(g(f(x)))=\sin(\sin(x^2))$
• $f(g(g(f(x))))=\left(\sin(\sin(x^2))\right)^2$

Therefore, $(f\circ g\circ g\circ f)(x)=\sin^2(\sin(x^2))$ and $(g\circ g\circ f)(x)=\sin(\sin(x^2)).$

So the equation becomes $\sin^2(\sin(x^2))=\sin(\sin(x^2)).$

Solve the equation

Let $y=\sin(\sin(x^2)).$ Then the equation is $y^2=y$ $\Rightarrow y(y-1)=0$ $\Rightarrow y=0 \quad \text{or} \quad y=1.$

Check possible values of $\sin(\sin(x^2))$

Since $\sin(x^2)\in[-1,1]$, we have $\sin(\sin(x^2))=\sin t \quad \text{for some } t\in[-1,1].$

Now for $t\in[-1,1]$, the value $\sin t$ lies in $[-\sin 1,\sin 1]$. In particular, $\sin(\sin(x^2))<1$ for every real $x$.

Hence the case $\sin(\sin(x^2))=1$ is impossible.

So only $\sin(\sin(x^2))=0$ is possible.

Solve $\sin(\sin(x^2))=0$

We know $\sin z=0 \iff z=n\pi,\quad n\in\mathbb Z.$

Here $z=\sin(x^2)$, and since $\sin(x^2)\in[-1,1],$ the only multiple of $\pi$ in the interval $[-1,1]$ is $0$.

Therefore, $\sin(x^2)=0.$

Solve $\sin(x^2)=0$

$\sin(x^2)=0 \iff x^2=n\pi,\quad n\in\{0,1,2,\dots\}$

because $x^2\ge 0$.

Thus, $x=\pm\sqrt{n\pi}, \qquad n\in\{0,1,2,\dots\}.$

Note that for $n=0$, this gives $x=0$.

Match with options

This is exactly Option A: $\pm\sqrt{n\pi},\; n\in\{0,1,2,\dots\}.$

Verification with stored answer

Stored correct answer: A

Our derived answer: A

So they agree.