We are given:

\[ \sin(4x) = \frac{1}{2}, \quad x \in (-9\pi,\ 3\pi) \]

Step 1: General solutions for \( \sin(θ) = \frac{1}{2} \)

\[ θ = \frac{\pi}{6} + 2n\pi \quad \text{or} \quad θ = \frac{5\pi}{6} + 2n\pi \]

Let \( θ = 4x \), so we get:

• \( x = \frac{\pi}{24} + \frac{n\pi}{2} \)
• \( x = \frac{5\pi}{24} + \frac{n\pi}{2} \)

✅ Step 2: Count how many such \( x \) fall in the interval \( (-9\pi, 3\pi) \)

By checking all possible \( n \) values, we find:

• For \( x = \frac{\pi}{24} + \frac{n\pi}{2} \): 24 valid values
• For \( x = \frac{5\pi}{24} + \frac{n\pi}{2} \): 24 valid values

? Total distinct values = 24 + 24 = 48

✅ Final Answer: $\boxed{48}$

Given: $$x^2 + 2y^2 = 3 \text{ and } x > y \text{ with } x, y \in \mathbb{Z}$$

Solutions to the equation are: $$\{(1,1), (1,-1), (-1,1), (-1,-1)\}$$

Among them, only \( (1, -1) \) satisfies \( x > y \).

Answer: $$\boxed{1}$$

? Maximum Students Passing All Three Exams

Given:

• Total students = 50
• \( |M| = 37 \), \( |P| = 24 \), \( |C| = 43 \)
• \( |M \cap P| \leq 19 \), \( |M \cap C| \leq 29 \), \( |P \cap C| \leq 20 \)

We use the inclusion-exclusion principle:

\[ |M \cup P \cup C| = |M| + |P| + |C| - |M \cap P| - |M \cap C| - |P \cap C| + |M \cap P \cap C| \]

Let \( x = |M \cap P \cap C| \). Then:

\[ 50 \geq 37 + 24 + 43 - 19 - 29 - 20 + x \Rightarrow 50 \geq 36 + x \Rightarrow x \leq 14 \]

✅ Final Answer: \(\boxed{14}\)