If the equation `sin theta(sin theta + 2 cos theta )=a` has a real solution then a belongs to the interval

Options

A. `[1/2,infty)`
B. `(-infty,-1/2]`
C. `[-1,1]`
D. `[-1/2,1/2]`

Answer

To determine the interval of \( a \) such that the equation \( \sin \theta (\sin \theta + 2 \cos \theta) = a \) has a real solution, we will analyze the expression step by step.

### Step 1: Rewrite the equation
We start with the equation:
\[ \sin \theta (\sin \theta + 2 \cos \theta) = a \]
This can be rewritten as:
\[ \sin^2 \theta + 2 \sin \theta \cos \theta = a \]

### Step 2: Use trigonometric identities
We can use the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \) to simplify the equation:
\[ \sin^2 \theta + \sin 2\theta = a \]

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

### Step 4: Analyze the expression
Let \( x = 2\theta \). Then the equation becomes:
\[ \cos x - 2 \sin x = 1 - 2a \]
We can express this in the form \( R \sin(x + \phi) = k \), where \( R = \sqrt{1^2 + (-2)^2} = \sqrt{5} \) and \( \tan \phi = -2 \).

### Step 5: Determine the range of \( k \)
The maximum and minimum values of \( R \sin(x + \phi) \) are \( R \) and \( -R \):
\[ -\sqrt{5} \leq \cos x - 2 \sin x \leq \sqrt{5} \]
Thus, we have:
\[ -\sqrt{5} \leq 1 - 2a \leq \sqrt{5} \]

### Step 6: Solve for \( a \)
Now we solve the inequalities:
1. From \( 1 - 2a \leq \sqrt{5} \):
\[ -2a \leq \sqrt{5} - 1 \implies a \geq \frac{1 - \sqrt{5}}{2} \]
2. From \( -\sqrt{5} \leq 1 - 2a \):
\[ -\sqrt{5} - 1 \leq -2a \implies 2a \leq 1 + \sqrt{5} \implies a \leq \frac{1 + \sqrt{5}}{2} \]

### Step 7: Combine the results
Combining the two inequalities gives:
\[ \frac{1 - \sqrt{5}}{2} \leq a \leq \frac{1 + \sqrt{5}}{2} \]

### Final Result
Thus, the interval for \( a \) such that the equation has a real solution is:
\[ \left[\frac{1 - \sqrt{5}}{2}, \frac{1 + \sqrt{5}}{2}\right] \]