The number of solution of `sin x+sin 2x+sin 3x`
`=cos x +cos 2x+cos 3x, 0 le x le 2pi`, is

Options

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

Explanation

To find the number of solutions for the equation \( \sin x + \sin 2x + \sin 3x = \cos x + \cos 2x + \cos 3x \) in the interval \( 0 \leq x \leq 2\pi \), we can follow these steps:

### Step 1: Rearranging the Equation
We start with the equation:
\[ \sin x + \sin 2x + \sin 3x - \cos x - \cos 2x - \cos 3x = 0 \]

### Step 2: Grouping Sine and Cosine Terms
We can rewrite the equation as:
\[ \sin x + \sin 2x + \sin 3x = \cos x + \cos 2x + \cos 3x \]
This can be rearranged to:
\[ \sin x + \sin 2x + \sin 3x - \cos x - \cos 2x - \cos 3x = 0 \]

### Step 3: Using Trigonometric Identities
Using the identity \( \sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right) \) and similar for cosine, we can simplify the terms. However, for this problem, we will directly analyze the equation.

### Step 4: Setting Up for Solutions
We can express the equation in terms of tangent:
\[ \tan 2x = 1 \]
This implies:
\[ 2x = n\pi + \frac{\pi}{4} \quad \text{for } n \in \mathbb{Z} \]
Thus:
\[ x = \frac{n\pi}{2} + \frac{\pi}{8} \]

### Step 5: Finding Solutions in the Interval
Now, we need to find the values of \( n \) such that \( 0 \leq x \leq 2\pi \):
1. For \( n = 0 \):
\[ x = \frac{\pi}{8} \]
2. For \( n = 1 \):
\[ x = \frac{3\pi}{8} \]
3. For \( n = 2 \):
\[ x = \frac{5\pi}{8} \]
4. For \( n = 3 \):
\[ x = \frac{7\pi}{8} \]
5. For \( n = 4 \):
\[ x = \frac{9\pi}{8} \]
6. For \( n = 5 \):
\[ x = \frac{11\pi}{8} \]
7. For \( n = 6 \):
\[ x = \frac{13\pi}{8} \]
8. For \( n = 7 \):
\[ x = \frac{15\pi}{8} \]
9. For \( n = 8 \):
\[ x = \frac{17\pi}{8} \]
10. For \( n = 9 \):
\[ x = \frac{19\pi}{8} \]

### Step 6: Counting Valid Solutions
Now we check which of these values fall within the interval \( [0, 2\pi] \):
- Valid solutions are \( \frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{7\pi}{8}, \frac{9\pi}{8}, \frac{11\pi}{8}, \frac{13\pi}{8}, \frac{15\pi}{8} \).

### Conclusion
Thus, the total number of solutions in the interval \( [0, 2\pi] \) is **8**.