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

Given, f(x) = x² for all x ∈ R
g(x) = sin x for all x ∈ R
Now, (f∘g∘g∘f)(x) = f(g(g(f(x)))) = f(g(g(x²))) = f(g(sin x²)) = f(sin(sin x²))
(f∘g∘g∘f)(x) = (sin(sin x²))²
Now, (g∘g∘f)(x) = g(g(f(x))) = g(g(x²)) = g(sin x²)
(g∘g∘f)(x) = sin(sin x²)
Given , (f∘g∘g∘f)(x) = (g∘g∘f)(x)
So, (sin(sin x²))² = sin(sin x²)
(sin(sin x²))² - sin(sin x²) = 0
sin(sin x²)(sin(sin x²) - 1) = 0
So, sin(sin x²) = 0 or sin(sin x²) = 1
⇒ sin x² = π/2 (it is not possible)
sin x² = 0 implies x² = nπ or x = ±√nπ, n ∈ 0, 1, 2, 3…
x = ±√nπ, n ∈ 0, 1, 2, 3…