If `sin theta+sin2theta+sin3theta=sin alpha` and `cos theta+cos 2theta+cos3theta=cos alpha`, then `theta` is equal to

Explanation

To solve the problem, we need to find the value of \( \theta \) given the equations:

1. \( \sin \theta + \sin 2\theta + \sin 3\theta = \sin \alpha \)
2. \( \cos \theta + \cos 2\theta + \cos 3\theta = \cos \alpha \)

### Step-by-step Solution:

**Step 1: Rewrite the sine equation using the sum-to-product identities.**

We can use the identity for the sum of sines:
\[ \sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right) \]

Applying this to \( \sin \theta + \sin 3\theta \):
\[ \sin \theta + \sin 3\theta = 2 \sin \left( \frac{\theta + 3\theta}{2} \right) \cos \left( \frac{3\theta - \theta}{2} \right) = 2 \sin (2\theta) \cos (\theta) \]

Thus, we can rewrite the first equation as:
\[ 2 \sin (2\theta) \cos (\theta) + \sin (2\theta) = \sin \alpha \]
Factoring out \( \sin (2\theta) \):
\[ \sin (2\theta) (2 \cos (\theta) + 1) = \sin \alpha \tag{1} \]

**Step 2: Rewrite the cosine equation using the sum-to-product identities.**

Using the identity for the sum of cosines:
\[ \cos A + \cos B = 2 \cos \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right) \]

Applying this to \( \cos \theta + \cos 3\theta \):
\[ \cos \theta + \cos 3\theta = 2 \cos \left( \frac{\theta + 3\theta}{2} \right) \cos \left( \frac{3\theta - \theta}{2} \right) = 2 \cos (2\theta) \cos (\theta) \]

Thus, we can rewrite the second equation as:
\[ 2 \cos (2\theta) \cos (\theta) + \cos (2\theta) = \cos \alpha \]
Factoring out \( \cos (2\theta) \):
\[ \cos (2\theta) (2 \cos (\theta) + 1) = \cos \alpha \tag{2} \]

**Step 3: Divide Equation (1) by Equation (2).**

We have:
\[ \frac{\sin (2\theta) (2 \cos (\theta) + 1)}{\cos (2\theta) (2 \cos (\theta) + 1)} = \frac{\sin \alpha}{\cos \alpha} \]
Assuming \( 2 \cos (\theta) + 1 \neq 0 \), we can cancel it:
\[ \frac{\sin (2\theta)}{\cos (2\theta)} = \tan \alpha \]
Thus,
\[ \tan (2\theta) = \tan \alpha \]

**Step 4: Solve for \( \theta \).**

From \( \tan (2\theta) = \tan \alpha \), we can conclude:
\[ 2\theta = \alpha + n\pi \quad (n \in \mathbb{Z}) \]
Thus,
\[ \theta = \frac{\alpha}{2} + \frac{n\pi}{2} \]

**Final Answer:**
The value of \( \theta \) is:
\[ \theta = \frac{\alpha}{2} + \frac{n\pi}{2} \]