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

Let f(x) = x² and g(x) = sin x for all x ∈ R. Then the set of all x satisfying (f ° g ° g ° f)(x) = (g ° g ° f)(x), where (f ° g)(x) = f(g(x)), is

$f(x)=x^{2},\quad g(x)=\sin x,\quad x\in\mathbb{R}$

$\text{To find: } x \text{ such that } (f\circ g\circ g\circ f)(x)=(g\circ g\circ f)(x)$

$f(x)=x^2 \;\Rightarrow\; g(f(x))=\sin(x^2)$

$g(g(f(x)))=\sin(\sin(x^2))$

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

$\left[\sin(\sin(x^2))\right]^2=\sin(\sin(x^2))$

$t=\sin(\sin(x^2)) \;\Rightarrow\; t^2=t$

$t(t-1)=0 \;\Rightarrow\; t=0 \ \text{or}\ t=1$

$\text{Plan: solve } \sin(\sin(x^2))=0 \ \text{and}\ \sin(\sin(x^2))=1$

$\sin(\sin(x^2))=0 \;\Rightarrow\; \sin(x^2)=0$

$x^2=n\pi,\ n\in\{0,1,2,\dots\} \;\Rightarrow\; x=\pm\sqrt{n\pi}$

$t=1 \;\Rightarrow\; \sin(x^2)=\dfrac{\pi}{2}+2m\pi > 1 \ \text{— impossible}$

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

$n=0:\ x=0 \text{ works} \;\Rightarrow\; \text{option }(1)$

$x=\pm\sqrt{n\pi},\ n\in\{0,1,2,\dots\} \;\Rightarrow\; \text{option }(1)$