The set of values of 'a' for which the equation `"sin" x ("sin"x +"cos" x) = a` has real solutions, is

Options

A. `[1-sqrt(2), 1+sqrt(2)]`
B. `[2-sqrt(3), 2+sqrt(3)]`
C. `[0, 2+sqrt(3)]`
D. `[(1-sqrt(2))/(2), (1+sqrt(2))/(2)]`

Explanation

To solve the equation \( \sin x (\sin x + \cos x) = a \) for the set of values of \( a \) for which it has real solutions, we can follow these steps:

### Step 1: Rewrite the equation
Start with the given equation:
\[ \sin x (\sin x + \cos x) = a \]
Expanding the left-hand side gives:
\[ \sin^2 x + \sin x \cos x = a \]

### Step 2: Multiply by 2
To facilitate the use of trigonometric identities, multiply both sides by 2:
\[ 2\sin^2 x + 2\sin x \cos x = 2a \]

### Step 3: Use trigonometric identities
Recall the double angle identities:
\[ 2\sin^2 x = 1 - \cos 2x \quad \text{and} \quad 2\sin x \cos x = \sin 2x \]
Substituting these identities into the equation gives:
\[ 1 - \cos 2x + \sin 2x = 2a \]
Rearranging this, we have:
\[ \sin 2x - \cos 2x = 2a - 1 \]

### Step 4: Express in terms of a single sine function
To express the left-hand side in a single sine function, we can multiply and divide by \( \sqrt{2} \):
\[ \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin 2x - \frac{1}{\sqrt{2}} \cos 2x \right) = 2a - 1 \]
This can be rewritten using the sine angle subtraction identity:
\[ \sqrt{2} \sin \left( 2x - \frac{\pi}{4} \right) = 2a - 1 \]

### Step 5: Determine the range of the sine function
The sine function ranges from -1 to 1, so:
\[ -1 \leq \sin \left( 2x - \frac{\pi}{4} \right) \leq 1 \]
This implies:
\[ -\sqrt{2} \leq 2a - 1 \leq \sqrt{2} \]

### Step 6: Solve the inequalities
Adding 1 to all parts of the inequality:
\[ 1 - \sqrt{2} \leq 2a \leq 1 + \sqrt{2} \]
Dividing the entire inequality by 2 gives:
\[ \frac{1 - \sqrt{2}}{2} \leq a \leq \frac{1 + \sqrt{2}}{2} \]

### Step 7: Final answer
Thus, the set of values of \( a \) for which the equation has real solutions is:
\[ a \in \left[ \frac{1 - \sqrt{2}}{2}, \frac{1 + \sqrt{2}}{2} \right] \]