Solve : `sin 2theta+sin 4theta+sin6 theta=0,(-180^(@)le theta<= 180^(@))`

Answer

To solve the equation \( \sin 2\theta + \sin 4\theta + \sin 6\theta = 0 \) for \( -180^\circ \leq \theta \leq 180^\circ \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \sin 2\theta + \sin 4\theta + \sin 6\theta = 0 \]

### Step 2: Group terms
We can rearrange the equation:
\[ \sin 6\theta + \sin 4\theta + \sin 2\theta = 0 \]

### Step 3: Use the sine addition formula
We can apply the sine addition formula:
\[ \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]
Let \( A = 6\theta \) and \( B = 2\theta \). Then:
\[ \sin 6\theta + \sin 2\theta = 2 \sin\left(4\theta\right) \cos\left(2\theta\right) \]
Thus, we can rewrite the equation as:
\[ 2 \sin 4\theta \cos 2\theta + \sin 4\theta = 0 \]

### Step 4: Factor out \( \sin 4\theta \)
Factoring out \( \sin 4\theta \):
\[ \sin 4\theta (2 \cos 2\theta + 1) = 0 \]

### Step 5: Set each factor to zero
Now we have two cases to consider:

**Case 1:**
\[ \sin 4\theta = 0 \]
This implies:
\[ 4\theta = n\pi \quad \Rightarrow \quad \theta = \frac{n\pi}{4} \]

**Case 2:**
\[ 2 \cos 2\theta + 1 = 0 \]
This simplifies to:
\[ \cos 2\theta = -\frac{1}{2} \]
This implies:
\[ 2\theta = (2n + 1)\frac{\pi}{3} \quad \Rightarrow \quad \theta = \frac{(2n + 1)\pi}{6} \]

### Step 6: Find solutions within the given range
Now we need to find the values of \( n \) that keep \( \theta \) within the range \( -180^\circ \leq \theta \leq 180^\circ \).

**For Case 1:**
\[ \theta = \frac{n\pi}{4} \]
- For \( n = -7 \): \( \theta = -\frac{7\pi}{4} \) (not in range)
- For \( n = -6 \): \( \theta = -\frac{3\pi}{2} \) (not in range)
- For \( n = -5 \): \( \theta = -\frac{5\pi}{4} \) (valid)
- For \( n = -4 \): \( \theta = -\pi \) (valid)
- For \( n = -3 \): \( \theta = -\frac{3\pi}{4} \) (valid)
- For \( n = -2 \): \( \theta = -\frac{\pi}{2} \) (valid)
- For \( n = -1 \): \( \theta = -\frac{\pi}{4} \) (valid)
- For \( n = 0 \): \( \theta = 0 \) (valid)
- For \( n = 1 \): \( \theta = \frac{\pi}{4} \) (valid)
- For \( n = 2 \): \( \theta = \frac{\pi}{2} \) (valid)
- For \( n = 3 \): \( \theta = \frac{3\pi}{4} \) (valid)
- For \( n = 4 \): \( \theta = \pi \) (valid)
- For \( n = 5 \): \( \theta = \frac{5\pi}{4} \) (not in range)
- For \( n = 6 \): \( \theta = \frac{3\pi}{2} \) (not in range)

**For Case 2:**
\[ \theta = \frac{(2n + 1)\pi}{6} \]
- For \( n = -6 \): \( \theta = -\frac{5\pi}{6} \) (valid)
- For \( n = -5 \): \( \theta = -\frac{3\pi}{6} = -\frac{\pi}{2} \) (valid)
- For \( n = -4 \): \( \theta = -\frac{\pi}{6} \) (valid)
- For \( n = -3 \): \( \theta = \frac{\pi}{6} \) (valid)
- For \( n = -2 \): \( \theta = \frac{3\pi}{6} = \frac{\pi}{2} \) (valid)
- For \( n = -1 \): \( \theta = \frac{5\pi}{6} \) (valid)
- For \( n = 0 \): \( \theta = \frac{7\pi}{6} \) (not in range)

### Final Solutions
The valid solutions for \( \theta \) are:
- From Case 1: \( -\frac{5\pi}{4}, -\pi, -\frac{3\pi}{4}, -\frac{\pi}{2}, -\frac{\pi}{4}, 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi \)
- From Case 2: \( -\frac{5\pi}{6}, -\frac{\pi}{2}, -\frac{\pi}{6}, \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6} \)