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
We start with the given equation:
\[ \sin^3 x \cos 3x + \sin 3x \cos^3 x = 0 \]
This can be factored as:
\[ \sin 3x \cos 3x + \sin^3 x \cos 3x + \sin 3x \cos^3 x = 0 \]
Rearranging gives:
\[ \sin 3x \cos 3x + \sin^3 x \cos 3x + \sin 3x \cos^3 x = 0 \]
We can factor out \(\sin 3x \cos 3x\):
\[ \sin 3x \cos 3x (1 + \tan^3 x) = 0 \]

### Step 2: Set each factor to zero
Now we have two factors:
1. \( \sin 3x = 0 \)
2. \( 1 + \tan^3 x = 0 \)

### Step 3: Solve \( \sin 3x = 0 \)
The solutions to \( \sin 3x = 0 \) occur when:
\[ 3x = n\pi \quad \text{for } n \in \mathbb{Z} \]
Thus,
\[ x = \frac{n\pi}{3} \]
We need to find \( n \) such that \( x \) is in the interval \([0, \pi]\):
- For \( n = 0 \): \( x = 0 \)
- For \( n = 1 \): \( x = \frac{\pi}{3} \)
- For \( n = 2 \): \( x = \frac{2\pi}{3} \)
- For \( n = 3 \): \( x = \pi \)

This gives us 4 solutions from this factor.

### Step 4: Solve \( 1 + \tan^3 x = 0 \)
This simplifies to:
\[ \tan^3 x = -1 \quad \Rightarrow \quad \tan x = -1 \]
The solutions for \( \tan x = -1 \) in the interval \([0, \pi]\) occur at:
\[ x = \frac{3\pi}{4} \]
This gives us 1 additional solution.

### Step 5: Count the total solutions
Adding the solutions from both factors:
- From \( \sin 3x = 0 \): \( 0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi \) (4 solutions)
- From \( \tan x = -1 \): \( \frac{3\pi}{4} \) (1 solution)

Thus, the total number of solutions in the interval \([0, \pi]\) is:
\[ 4 + 1 = 5 \]

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

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