Number of solutions of the equation
`sin 5x cdot cos 3x = sin 6x cdot cos 2x`,
In the interval `[0, pi]` is

Options

A. 3
B. 4
C. 5
D. none of these

Answer

To find the number of solutions for the equation

\[ \sin(5x) \cdot \cos(3x) = \sin(6x) \cdot \cos(2x) \]

in the interval \([0, \pi]\), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:

\[ \sin(5x) \cdot \cos(3x) = \sin(6x) \cdot \cos(2x) \]

### Step 2: Use the product-to-sum identities
We can apply the product-to-sum identities, which state:

\[ \sin A \cos B = \frac{1}{2} [\sin(A + B) + \sin(A - B)] \]

Applying this to both sides:

\[ \sin(5x) \cdot \cos(3x) = \frac{1}{2} [\sin(8x) + \sin(2x)] \]
\[ \sin(6x) \cdot \cos(2x) = \frac{1}{2} [\sin(8x) + \sin(4x)] \]

### Step 3: Set the equations equal
Now we have:

\[ \frac{1}{2} [\sin(8x) + \sin(2x)] = \frac{1}{2} [\sin(8x) + \sin(4x)] \]

Multiplying through by 2 to eliminate the fraction:

\[ \sin(8x) + \sin(2x) = \sin(8x) + \sin(4x) \]

### Step 4: Simplify the equation
Subtract \(\sin(8x)\) from both sides:

\[ \sin(2x) = \sin(4x) \]

### Step 5: Use the sine difference identity
Using the identity \(\sin A = \sin B\), we can write:

\[ \sin(2x) - \sin(4x) = 0 \]

Using the sine difference formula:

\[ 2 \sin\left(\frac{2x - 4x}{2}\right) \cos\left(\frac{2x + 4x}{2}\right) = 0 \]

This simplifies to:

\[ 2 \sin(-x) \cos(3x) = 0 \]

### Step 6: Set each factor to zero
This gives us two cases to consider:

1. \(\sin(-x) = 0\)
2. \(\cos(3x) = 0\)

### Step 7: Solve \(\sin(-x) = 0\)
The general solution for \(\sin(-x) = 0\) is:

\[ -x = n\pi \implies x = -n\pi \]

In the interval \([0, \pi]\), the only solution is:

\[ x = 0 \quad (n = 0) \]

### Step 8: Solve \(\cos(3x) = 0\)
The general solution for \(\cos(3x) = 0\) is:

\[ 3x = \frac{(2n + 1)\pi}{2} \implies x = \frac{(2n + 1)\pi}{6} \]

### Step 9: Find valid \(n\) values
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 \(n = 1\):
\[ x = \frac{3\pi}{6} = \frac{\pi}{2} \]

- For \(n = 2\):
\[ x = \frac{5\pi}{6} \]

- For \(n = 3\):
\[ x = \frac{7\pi}{6} \quad (\text{not in } [0, \pi]) \]

### Step 10: Count the solutions
The solutions we found in the interval \([0, \pi]\) are:

1. \(x = 0\)
2. \(x = \frac{\pi}{6}\)
3. \(x = \frac{\pi}{2}\)
4. \(x = \frac{5\pi}{6}\)

Thus, the total number of solutions is:

\[ \text{Number of solutions} = 4 \]

### Final Answer
The number of solutions of the equation \(\sin(5x) \cdot \cos(3x) = \sin(6x) \cdot \cos(2x)\) in the interval \([0, \pi]\) is **4**.

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