Let f(x) = x^2 and g(x) = sinx for all x belongs to R. Then find the set of all x satisfying (fogogof)(x) = (gogof)(x),
where (fog)(x) = f(g(x)).

To solve the equation \((f \circ g \circ g \circ f)(x) = (g \circ g \circ f)(x)\), we first need to understand what the functions \(f\) and \(g\) represent and how to compose them. Here, \(f(x) = x^2\) and \(g(x) = \sin x\). The notation \((f \circ g)(x)\) means we apply \(g\) first and then \(f\), which can be expressed as \(f(g(x))\). Let's break this down step by step.

Step 1: Calculate \(f \circ g\)

Starting with the composition \(f(g(x))\):

• We know \(g(x) = \sin x\).
• Thus, \(f(g(x)) = f(\sin x) = (\sin x)^2\).

So, we have:

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

Step 2: Calculate \(g \circ f\)

Next, we compute \(g(f(x))\):

• Here, \(f(x) = x^2\).
• Therefore, \(g(f(x)) = g(x^2) = \sin(x^2)\).

Thus, we find:

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

Step 3: Calculate \(f \circ g \circ g\)

Now, we need to find \((f \circ g \circ g)(x)\):

• First, compute \(g(g(x)) = g(\sin x) = \sin(\sin x)\).
• Then, apply \(f\): \(f(g(g(x))) = f(\sin(\sin x)) = (\sin(\sin x))^2\).

So, we have:

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

Step 4: Calculate \(g \circ g \circ f\)

Next, we compute \((g \circ g \circ f)(x)\):

• First, find \(g(f(x)) = g(x^2) = \sin(x^2)\).
• Then, apply \(g\) again: \(g(g(f(x))) = g(\sin(x^2)) = \sin(\sin(x^2))\).

Thus, we have:

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

Step 5: Set the equations equal

Now we set the two results equal to each other:

\((\sin(\sin x))^2 = \sin(\sin(x^2))\

Step 6: Analyze the equation

This equation can be tricky to solve directly, but we can analyze it:

• The left side, \((\sin(\sin x))^2\), is always non-negative since it is a square.
• The right side, \(\sin(\sin(x^2))\), can take values between -1 and 1.

For the equality to hold, \(\sin(\sin(x^2))\) must also be non-negative, which means:

\(\sin(\sin(x^2)) \geq 0\)

Step 7: Finding solutions

To find the solutions, we need to consider when \(\sin(\sin x) = 0\) or \(\sin(\sin x) = 1\). The sine function is zero at integer multiples of \(\pi\), so:

• \(\sin x = n\pi\) for \(n \in \mathbb{Z}\) leads to \(x = \arcsin(n\pi)\), but we must restrict \(n\) such that \(-1 \leq n\pi \leq 1\).

Thus, the only feasible values for \(n\) are 0, leading to \(x = 0\) as a solution.

Final Thoughts

In conclusion, the set of all \(x\) satisfying the equation \((f \circ g \circ g \circ f)(x) = (g \circ g \circ f)(x)\) is:

\(\{0\}\)