The number of solution of `sinx+sin3x+sin5x=0` in the interval `{:[(pi)/(2)","(3pi)/(2)]:}` is

Options

A. 2
B. 3
C. 4
D. 5

Explanation

To find the number of solutions of the equation \( \sin x + \sin 3x + \sin 5x = 0 \) in the interval \( \left[\frac{\pi}{2}, \frac{3\pi}{2}\right] \), we can follow these steps:

### Step 1: Group the Sine Terms
We can group the sine terms as follows:
\[ \sin x + \sin 5x + \sin 3x = 0 \]
We can combine \( \sin x \) and \( \sin 5x \) using the sine addition formula.

### Step 2: Use the Sine Addition Formula
Recall the formula:
\[ \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]
Let \( A = x \) and \( B = 5x \):
\[ \sin x + \sin 5x = 2 \sin\left(\frac{x + 5x}{2}\right) \cos\left(\frac{x - 5x}{2}\right) = 2 \sin(3x) \cos(2x) \]
Thus, we can rewrite the equation as:
\[ 2 \sin(3x) \cos(2x) + \sin(3x) = 0 \]

### Step 3: Factor the Equation
Factoring out \( \sin(3x) \):
\[ \sin(3x)(2 \cos(2x) + 1) = 0 \]

### Step 4: Solve Each Factor
We have two cases to solve:
1. \( \sin(3x) = 0 \)
2. \( 2 \cos(2x) + 1 = 0 \)

**Case 1: Solve \( \sin(3x) = 0 \)**
\[ 3x = n\pi \quad \Rightarrow \quad x = \frac{n\pi}{3} \]
We need to find \( n \) such that \( \frac{\pi}{2} \leq \frac{n\pi}{3} \leq \frac{3\pi}{2} \).

Multiplying through by \( 3 \):
\[ \frac{3\pi}{2} \leq n\pi \leq \frac{9\pi}{2} \]
Dividing through by \( \pi \):
\[ \frac{3}{2} \leq n \leq \frac{9}{2} \]
Thus, \( n = 2, 3, 4 \) gives us:
- For \( n = 2 \): \( x = \frac{2\pi}{3} \)
- For \( n = 3 \): \( x = \pi \)
- For \( n = 4 \): \( x = \frac{4\pi}{3} \)

**Case 2: Solve \( 2 \cos(2x) + 1 = 0 \)**
\[ \cos(2x) = -\frac{1}{2} \]
The general solutions for \( \cos(2x) = -\frac{1}{2} \) are:
\[ 2x = \frac{2\pi}{3} + 2k\pi \quad \text{or} \quad 2x = \frac{4\pi}{3} + 2k\pi \]
Thus:
\[ x = \frac{\pi}{3} + k\pi \quad \text{or} \quad x = \frac{2\pi}{3} + k\pi \]
For \( k = 0 \):
- \( x = \frac{\pi}{3} \) (not in the interval)
- \( x = \frac{2\pi}{3} \) (in the interval)

For \( k = 1 \):
- \( x = \frac{4\pi}{3} \) (in the interval)

### Step 5: Count the Solutions
From both cases, the valid solutions in the interval \( \left[\frac{\pi}{2}, \frac{3\pi}{2}\right] \) are:
1. \( x = \frac{2\pi}{3} \)
2. \( x = \pi \)
3. \( x = \frac{4\pi}{3} \)

Thus, the total number of solutions is **3**.

### Final Answer
The number of solutions of \( \sin x + \sin 3x + \sin 5x = 0 \) in the interval \( \left[\frac{\pi}{2}, \frac{3\pi}{2}\right] \) is **3**.