If the equation `"sin" theta ("sin" theta + 2 "cos" theta) = a` has a real solution, then the shortest interval containing 'a', is

Options

A. `[(1-sqrt(5))/(2), (1+sqrt(5))/(2)]`
B. `((sqrt(5)-1)/(2), (sqrt(5)+1)/(2))`
C. `(-1//2, 1//2)`
D. none of these

Answer

To solve the equation \( \sin \theta (\sin \theta + 2 \cos \theta) = a \) and determine the shortest interval containing \( a \) for which the equation has a real solution, we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \sin \theta (\sin \theta + 2 \cos \theta) = a \]
Expanding this, we have:
\[ \sin^2 \theta + 2 \sin \theta \cos \theta = a \]

### Step 2: Use trigonometric identities
We know that \( 2 \sin \theta \cos \theta = \sin(2\theta) \). Thus, we can rewrite the equation as:
\[ \sin^2 \theta + \sin(2\theta) = a \]

### Step 3: Express \( \sin^2 \theta \) in terms of \( \cos(2\theta) \)
Using the identity \( \sin^2 \theta = \frac{1 - \cos(2\theta)}{2} \), we substitute:
\[ \frac{1 - \cos(2\theta)}{2} + \sin(2\theta) = a \]
Multiplying through by 2 to eliminate the fraction gives:
\[ 1 - \cos(2\theta) + 2\sin(2\theta) = 2a \]
Rearranging this, we have:
\[ -\cos(2\theta) + 2\sin(2\theta) = 2a - 1 \]

### Step 4: Set up the function
Let \( t = 2\theta \). Then we can express the equation as:
\[ 2\sin(t) - \cos(t) = 2a - 1 \]
This can be rewritten as:
\[ \sqrt{(2^2 + (-1)^2)} \left( \frac{2}{\sqrt{5}} \sin(t) - \frac{1}{\sqrt{5}} \cos(t) \right) = 2a - 1 \]
The magnitude of the left-hand side is \( \sqrt{5} \), so we have:
\[ -\sqrt{5} \leq 2a - 1 \leq \sqrt{5} \]

### Step 5: Solve for \( a \)
Adding 1 to all parts of the inequality gives:
\[ 1 - \sqrt{5} \leq 2a \leq 1 + \sqrt{5} \]
Dividing through by 2 results in:
\[ \frac{1 - \sqrt{5}}{2} \leq a \leq \frac{1 + \sqrt{5}}{2} \]

### Conclusion
Thus, the shortest interval containing \( a \) is:
\[ \left[ \frac{1 - \sqrt{5}}{2}, \frac{1 + \sqrt{5}}{2} \right] \]