The number of solutions of the equation `"sin" x = "cos" 3x " in " [0, pi]` is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To find the number of solutions for the equation \( \sin x = \cos 3x \) in the interval \( [0, \pi] \), we can follow these steps:

### Step 1: Rewrite the Equation
We start with the equation:
\[ \sin x = \cos 3x \]
Using the identity \( \cos 3x = \sin\left(\frac{\pi}{2} - 3x\right) \), we can rewrite the equation as:
\[ \sin x = \sin\left(\frac{\pi}{2} - 3x\right) \]

### Step 2: Set Up the General Solutions
The equation \( \sin A = \sin B \) implies:
\[ A = B + n\pi \quad \text{or} \quad A = \pi - B + n\pi, \quad n \in \mathbb{Z} \]
Applying this to our equation gives us two cases:
1. \( x = \frac{\pi}{2} - 3x + n\pi \)
2. \( x = \pi - \left(\frac{\pi}{2} - 3x\right) + n\pi \)

### Step 3: Solve the First Case
From the first case:
\[ x + 3x = \frac{\pi}{2} + n\pi \implies 4x = \frac{\pi}{2} + n\pi \implies x = \frac{\pi}{8} + \frac{n\pi}{4} \]

### Step 4: Determine Valid Values of \( n \)
We need to find values of \( n \) such that \( x \) is in the interval \( [0, \pi] \):
- For \( n = 0 \): \( x = \frac{\pi}{8} \)
- For \( n = 1 \): \( x = \frac{\pi}{8} + \frac{\pi}{4} = \frac{3\pi}{8} \)
- For \( n = 2 \): \( x = \frac{\pi}{8} + \frac{2\pi}{4} = \frac{5\pi}{8} \)
- For \( n = 3 \): \( x = \frac{\pi}{8} + \frac{3\pi}{4} = \frac{7\pi}{8} \)
- For \( n = 4 \): \( x = \frac{\pi}{8} + \frac{4\pi}{4} = \frac{9\pi}{8} \) (not valid since \( \frac{9\pi}{8} > \pi \))

Thus, from the first case, we have valid solutions:
\[ x = \frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{7\pi}{8} \]

### Step 5: Solve the Second Case
From the second case:
\[ x = \pi - \left(\frac{\pi}{2} - 3x\right) + n\pi \implies x = \pi - \frac{\pi}{2} + 3x + n\pi \implies -2x = -\frac{\pi}{2} + n\pi \implies 2x = \frac{\pi}{2} - n\pi \implies x = \frac{\pi}{4} - \frac{n\pi}{2} \]

### Step 6: Determine Valid Values of \( n \) for the Second Case
For this case, we also need \( x \) to be in \( [0, \pi] \):
- For \( n = 0 \): \( x = \frac{\pi}{4} \)
- For \( n = 1 \): \( x = \frac{\pi}{4} - \frac{\pi}{2} = -\frac{\pi}{4} \) (not valid)
- For \( n = -1 \): \( x = \frac{\pi}{4} + \frac{\pi}{2} = \frac{3\pi}{4} \)

Thus, from the second case, we have valid solutions:
\[ x = \frac{\pi}{4}, \frac{3\pi}{4} \]

### Step 7: Combine All Solutions
Combining the solutions from both cases, we have:
1. \( x = \frac{\pi}{8} \)
2. \( x = \frac{3\pi}{8} \)
3. \( x = \frac{5\pi}{8} \)
4. \( x = \frac{7\pi}{8} \)
5. \( x = \frac{\pi}{4} \)
6. \( x = \frac{3\pi}{4} \)

### Step 8: Count Unique Solutions
The unique solutions in the interval \( [0, \pi] \) are:
- \( \frac{\pi}{8} \)
- \( \frac{3\pi}{8} \)
- \( \frac{\pi}{4} \)
- \( \frac{5\pi}{8} \)
- \( \frac{3\pi}{4} \)
- \( \frac{7\pi}{8} \)

Thus, the total number of solutions is **6**.

### Final Answer
The number of solutions of the equation \( \sin x = \cos 3x \) in the interval \( [0, \pi] \) is **6**.