The equation `sin x ( sin x + cos x ) = k` has real solution if and only if k is a real number such that

Options

A. `0 le k le ( 1+ sqrt( 2))/( 2)`
B. `2- sqrt( 3) le k le 2 + sqrt(3)`
C. `0 le k le2 - sqrt( 3)`
D. `( 1- sqrt( 2))/( 2) le k le ( 1+ sqrt( 2))/( 2)`

Answer

To find the range of \( k \) for which the equation \( \sin x (\sin x + \cos x) = k \) has real solutions, we will analyze the expression \( \sin x (\sin x + \cos x) \).

### Step 1: Expand the expression
Start by expanding the left-hand side:
\[ \sin x (\sin x + \cos x) = \sin^2 x + \sin x \cos x \]

### Step 2: Rewrite in terms of trigonometric identities
We can rewrite \( \sin x \cos x \) using the double angle identity:
\[ \sin x \cos x = \frac{1}{2} \sin 2x \]
Thus, we have:
\[ \sin^2 x + \sin x \cos x = \sin^2 x + \frac{1}{2} \sin 2x \]

### Step 3: Find the range of \( \sin^2 x \)
The range of \( \sin^2 x \) is from 0 to 1, since \( \sin x \) varies between -1 and 1.

### Step 4: Analyze the expression \( \sin^2 x + \frac{1}{2} \sin 2x \)
We need to find the maximum and minimum values of the expression \( \sin^2 x + \frac{1}{2} \sin 2x \).

### Step 5: Use trigonometric identities to find maximum and minimum
Using the identity \( \sin^2 x = \frac{1 - \cos 2x}{2} \), we can express:
\[ \sin^2 x + \frac{1}{2} \sin 2x = \frac{1 - \cos 2x}{2} + \frac{1}{2} \sin 2x \]
Let \( y = \sin 2x \). The maximum value of \( y \) is 1 and the minimum is -1.

### Step 6: Find the maximum and minimum values of the new expression
The expression can be rewritten as:
\[ \frac{1}{2} - \frac{1}{2} \cos 2x + \frac{1}{2} y \]
The maximum value occurs when both \( y \) and \( \cos 2x \) are at their respective maximum values, which gives:
\[ \text{Max} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2} \]
The minimum occurs when \( y = -1 \) and \( \cos 2x = 1 \):
\[ \text{Min} = \frac{1}{2} - \frac{1}{2} - \frac{1}{2} = -\frac{1}{2} \]

### Step 7: Determine the range of \( k \)
Thus, the range of \( k \) for which the equation has real solutions is:
\[ -\frac{1}{2} \leq k \leq \frac{3}{2} \]

### Final Result
The equation \( \sin x (\sin x + \cos x) = k \) has real solutions if and only if:
\[ k \in \left[-\frac{1}{2}, \frac{3}{2}\right] \]