The problem is asking to work with two functions:

$f(x) = 2x + \frac{3}{2}\pi \quad \text{and} \quad g(x) = 2\sin(x) + \sqrt{2}.$

Part (a)

We need to find the expression for the composition $(g \circ f)(x)$, which means we will substitute $f(x)$ into $g(x)$:

$(g \circ f)(x) = g(f(x)) = g\left(2x + \frac{3}{2}\pi\right).$

Substitute $f(x)$ into $g(x)$:

$g\left(2x + \frac{3}{2}\pi\right) = 2\sin\left(2x + \frac{3}{2}\pi\right) + \sqrt{2}.$

Now simplify the sine expression using the trigonometric identity $\sin\left(x + \frac{3}{2}\pi\right) = -\cos(x)$:

$\sin\left(2x + \frac{3}{2}\pi\right) = -\cos(2x).$

So:

$g\left(2x + \frac{3}{2}\pi\right) = 2(-\cos(2x)) + \sqrt{2} = -2\cos(2x) + \sqrt{2}.$

Thus, the expression for $(g \circ f)(x)$ is:

$(g \circ f)(x) = -2\cos(2x) + \sqrt{2}.$

Part (b)

The set $C_0(g \circ f)$ represents the values of $x$ for which $(g \circ f)(x) = 0$. Therefore, we need to solve the equation:

$-2\cos(2x) + \sqrt{2} = 0.$

Rearranging:

$-2\cos(2x) = -\sqrt{2},$ $\cos(2x) = \frac{\sqrt{2}}{2}.$

The general solution to $\cos(2x) = \frac{\sqrt{2}}{2}$ is:

$2x = \pm \frac{\pi}{4} + 2k\pi \quad \text{for} \; k \in \mathbb{Z}.$

Thus,

$x = \pm \frac{\pi}{8} + k\pi \quad \text{for} \; k \in \mathbb{Z}.$

Part (c)

Now, we need to find the intersection of $C_0(g \circ f)$ with the interval $[0, \pi]$, meaning we only take solutions from part (b) that lie within $[0, \pi]$.

Let's check the specific values for $x = \frac{\pi}{8} + k\pi$ and $x = -\frac{\pi}{8} + k\pi$:

For $k = 0$:

• $x = \frac{\pi}{8}$, which is in $[0, \pi]$,
• $x = -\frac{\pi}{8}$, which is not in $[0, \pi]$.

For $k = 1$:

• $x = \frac{\pi}{8} + \pi = \frac{9\pi}{8}$, which is greater than $\pi$, so it's not in $[0, \pi]$.

Thus, the only solution in $[0, \pi]$ is:

$x = \frac{\pi}{8}.$

Summary of answers:

• (a) $(g \circ f)(x) = -2\cos(2x) + \sqrt{2}$.
• (b) $C_0(g \circ f) = \left\{ x = \pm \frac{\pi}{8} + k\pi \mid k \in \mathbb{Z} \right\}$.
• (c) $C_0(g \circ f) \cap [0, \pi] = \left\{ \frac{\pi}{8} \right\}$.

Let me know if you need further clarification or additional details!

Here are 5 follow-up questions you could consider:

1. How do you derive the general solution for trigonometric equations like $\cos(2x) = \frac{\sqrt{2}}{2}$?
2. What are the steps to simplify trigonometric identities like $\sin(x + \frac{3}{2}\pi)$?
3. How would the result change if the interval in part (c) was expanded beyond $[0, \pi]$?
4. Can you explain how to determine the composition of two functions in general?
5. How does the range of $f(x)$ affect the domain of $g(f(x))$?

Tip: Always double-check the signs and periodicity when solving trigonometric equations, especially when they involve transformations like $\cos(2x)$.