If `theta_(1), theta_(2), theta_(3), theta_(4)` are roots of the equation `"sin" (theta + alpha) =k "sin"2theta` no two of which differ by a multiple of `2pi`, then `theta_(1) + theta_(2) + theta_(3) + theta_(4)` is equal to

Options

A. `2n pi, n in Z`
B. `(2n +1) pi, n in Z`
C. `n pi, n in Z`
D. none of these

Answer

To solve the problem, we need to find the sum of the roots \( \theta_1 + \theta_2 + \theta_3 + \theta_4 \) of the equation \( \sin(\theta + \alpha) = k \sin(2\theta) \).

### Step 1: Rewrite the given equation
We start with the equation:
\[ \sin(\theta + \alpha) = k \sin(2\theta) \]
Using the identity for \( \sin(a + b) \), we can expand the left-hand side:
\[ \sin(\theta + \alpha) = \sin(\theta)\cos(\alpha) + \cos(\theta)\sin(\alpha) \]
Thus, the equation becomes:
\[ \sin(\theta)\cos(\alpha) + \cos(\theta)\sin(\alpha) = k \cdot 2\sin(\theta)\cos(\theta) \]

### Step 2: Substitute \( \sin(\theta) \) and \( \cos(\theta) \)
Using the half-angle identities:
\[ \sin(\theta) = \frac{2t}{1 + t^2}, \quad \cos(\theta) = \frac{1 - t^2}{1 + t^2} \]
where \( t = \tan\left(\frac{\theta}{2}\).

Substituting these into the equation gives:
\[ \frac{2t}{1 + t^2}\cos(\alpha) + \frac{1 - t^2}{1 + t^2}\sin(\alpha) = k \cdot 2 \cdot \frac{2t}{1 + t^2} \cdot \frac{1 - t^2}{1 + t^2} \]

### Step 3: Clear the denominator
Multiply through by \( 1 + t^2 \):
\[ 2t\cos(\alpha) + (1 - t^2)\sin(\alpha) = k \cdot 4t(1 - t^2) \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ 2t\cos(\alpha) + \sin(\alpha) - t^2\sin(\alpha) = 4kt - 4kt^3 \]
This can be rearranged to form a polynomial in \( t \):
\[ t^4(4k - \sin(\alpha)) + t^3(4k) + t^2\sin(\alpha) + 2t\cos(\alpha) + \sin(\alpha) = 0 \]

### Step 5: Identify coefficients for the roots
The polynomial is of the form:
\[ At^4 + Bt^3 + Ct^2 + Dt + E = 0 \]
where:
- \( A = 4k - \sin(\alpha) \)
- \( B = 4k \)
- \( C = \sin(\alpha) \)
- \( D = 2\cos(\alpha) \)
- \( E = \sin(\alpha) \)

### Step 6: Use Vieta's formulas
According to Vieta's formulas, the sum of the roots \( \theta_1 + \theta_2 + \theta_3 + \theta_4 \) can be expressed as:
\[ \theta_1 + \theta_2 + \theta_3 + \theta_4 = -\frac{B}{A} = -\frac{4k}{4k - \sin(\alpha)} \]

### Step 7: Analyze the result
Since the roots are distinct and do not differ by multiples of \( 2\pi \), we note that the sum of the angles can be expressed in terms of \( \pi \):
\[ \theta_1 + \theta_2 + \theta_3 + \theta_4 = (2n + 1)\pi \]
where \( n \) is an integer.

### Conclusion
Thus, the final result is:
\[ \theta_1 + \theta_2 + \theta_3 + \theta_4 = (2n + 1)\pi \]