https://file.aiquickdraw.com/mathbot/file/1728522404971m5tbycu8.jpg

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 \(\cos(2x)\).

Sean f(x) = 2x + (3/2)π y g(x) = 2sen(x) + √2. (a) Hallar la expresión de (g o f)(x). (b) Hallar C_0(g o f). (c) Hallar C_0(g o f) ∩ [0, π].

Learn how to compose the functions f(x) = 2x + (3/2)π and g(x) = 2sin(x) + √2 and solve the corresponding trigonometric equations. This problem covers function composition, trigonometric identities, and solving equations involving sine and cosine. Suitable for early calculus students.

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