The equation `sin x(sinx+cos x) = k` has real solutions, where k is a real number. Then

Options

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

Answer

To solve the equation \( \sin x (\sin x + \cos x) = k \) for real solutions, we will follow these steps:

### Step 1: Expand the equation
Start by expanding the left side of the equation:
\[ \sin x (\sin x + \cos x) = \sin^2 x + \sin x \cos x \]
So, we can rewrite the equation as:
\[ \sin^2 x + \sin x \cos x = k \]

### Step 2: Use trigonometric identities
We can use the double angle identities to express the equation in a more manageable form. Recall that:
\[ \sin 2x = 2 \sin x \cos x \]
Thus, we can express \( \sin x \cos x \) as:
\[ \sin x \cos x = \frac{1}{2} \sin 2x \]
Substituting this back into our equation gives:
\[ \sin^2 x + \frac{1}{2} \sin 2x = k \]

### Step 3: Rewrite \( \sin^2 x \)
Using the identity \( \sin^2 x = \frac{1 - \cos 2x}{2} \), we can rewrite the equation:
\[ \frac{1 - \cos 2x}{2} + \frac{1}{2} \sin 2x = k \]
Multiplying through by 2 to eliminate the fraction:
\[ 1 - \cos 2x + \sin 2x = 2k \]
This simplifies to:
\[ \sin 2x - \cos 2x = 2k - 1 \]

### Step 4: Find the range of \( \sin 2x - \cos 2x \)
The expression \( \sin 2x - \cos 2x \) can be rewritten as:
\[ \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin 2x - \frac{1}{\sqrt{2}} \cos 2x \right) \]
This can be expressed as:
\[ \sqrt{2} \sin \left( 2x - \frac{\pi}{4} \right) \]
The maximum value of \( \sin \) function is 1 and the minimum value is -1. Therefore:
\[ -\sqrt{2} \leq \sin 2x - \cos 2x \leq \sqrt{2} \]

### Step 5: Set up the inequality for \( k \)
From the equation \( \sin 2x - \cos 2x = 2k - 1 \), we can set up the following inequalities:
\[ -\sqrt{2} \leq 2k - 1 \leq \sqrt{2} \]

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

### Conclusion
Thus, the values of \( k \) for which the equation has real solutions are given by:
\[ \frac{1 - \sqrt{2}}{2} \leq k \leq \frac{1 + \sqrt{2}}{2} \]