The number of solutions in the interval `[0, pi]` of the equation `sin^(3)x cos 3x+sin 3xcos^(3)x=0` is equal to

Options

A. 7
B. 6
C. 5
D. 4

Answer

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

### Step 1: Factor the equation
The given equation can be factored as:
\[ \sin^3 x \cos 3x + \sin 3x \cos^3 x = 0 \]
This can be rewritten as:
\[ \sin x \cos 3x (\sin^2 x + \sin 3x \cos^2 x) = 0 \]

### Step 2: Set each factor to zero
We have two factors to consider:
1. \( \sin x = 0 \)
2. \( \cos 3x = 0 \)
3. \( \sin^2 x + \sin 3x \cos^2 x = 0 \) (we will analyze this later)

### Step 3: Solve \( \sin x = 0 \)
The solutions to \( \sin x = 0 \) in the interval \([0, \pi]\) are:
\[ x = 0, \pi \]
Thus, there are 2 solutions from this factor.

### Step 4: Solve \( \cos 3x = 0 \)
The solutions to \( \cos 3x = 0 \) occur at:
\[ 3x = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]
This simplifies to:
\[ x = \frac{\pi}{6} + \frac{n\pi}{3} \]
Now, we need to find values of \(n\) such that \(x\) is in the interval \([0, \pi]\):
- For \(n = 0\): \(x = \frac{\pi}{6}\)
- For \(n = 1\): \(x = \frac{\pi}{2}\)
- For \(n = 2\): \(x = \frac{5\pi}{6}\)
- For \(n = 3\): \(x = \frac{3\pi}{2}\) (not in the interval)

Thus, we have 3 solutions from this factor: \(x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}\).

### Step 5: Analyze \( \sin^2 x + \sin 3x \cos^2 x = 0 \)
This equation can be rearranged to:
\[ \sin^2 x = -\sin 3x \cos^2 x \]
Since \(\sin^2 x\) is always non-negative, this equation has no solutions in the interval \([0, \pi]\).

### Step 6: Count total solutions
Now, we combine the solutions:
- From \( \sin x = 0 \): 2 solutions
- From \( \cos 3x = 0 \): 3 solutions
- From \( \sin^2 x + \sin 3x \cos^2 x = 0 \): 0 solutions

Total number of solutions in the interval \([0, \pi]\) is:
\[ 2 + 3 = 5 \]

### Final Answer
The number of solutions in the interval \([0, \pi]\) is **5**.