The values of a for which the equation `sqrtasinx-2cosx=sqrt2+sqrt(2-a)` has solutions are

Options

A. `p gt 0`
B. `p le 3`
C. `1 le p le 2`
D. `sqrt(5)-1 le p le 2`

Explanation

To solve the equation \( \sqrt{a} \sin x - 2 \cos x = \sqrt{2} + \sqrt{2 - a} \) for the values of \( a \) for which it has solutions, we will follow these steps:

### Step 1: Rewrite the Equation
We start with the equation:
\[ \sqrt{a} \sin x - 2 \cos x = \sqrt{2} + \sqrt{2 - a} \]
We can express this in the form \( A \sin x + B \cos x = C \), where:
- \( A = \sqrt{a} \)
- \( B = -2 \)
- \( C = \sqrt{2} + \sqrt{2 - a} \)

### Step 2: Determine the Range of the Left Side
The expression \( A \sin x + B \cos x \) can be represented as:
\[ R \sin(x + \phi) \]
where \( R = \sqrt{A^2 + B^2} = \sqrt{(\sqrt{a})^2 + (-2)^2} = \sqrt{a + 4} \).

The maximum value of \( R \sin(x + \phi) \) is \( R \), and the minimum value is \( -R \). Therefore, we have:
\[ -\sqrt{a + 4} \leq \sqrt{a} \sin x - 2 \cos x \leq \sqrt{a + 4} \]

### Step 3: Set Up Inequalities
For the equation to have solutions, the right side must fall within the range of the left side:
\[ -\sqrt{a + 4} \leq \sqrt{2} + \sqrt{2 - a} \leq \sqrt{a + 4} \]

### Step 4: Solve the First Inequality
Starting with the first inequality:
\[ \sqrt{2} + \sqrt{2 - a} \geq -\sqrt{a + 4} \]
Squaring both sides (noting that both sides are non-negative):
\[ (\sqrt{2} + \sqrt{2 - a})^2 \geq a + 4 \]
Expanding the left side:
\[ 2 + 2\sqrt{2(2 - a)} + (2 - a) \geq a + 4 \]
Simplifying:
\[ 2\sqrt{2(2 - a)} \geq 2a + 2 \]
Dividing by 2:
\[ \sqrt{2(2 - a)} \geq a + 1 \]
Squaring again:
\[ 2(2 - a) \geq (a + 1)^2 \]
Expanding:
\[ 4 - 2a \geq a^2 + 2a + 1 \]
Rearranging:
\[ 0 \geq a^2 + 4a - 3 \]
This is a quadratic inequality.

### Step 5: Find Roots of the Quadratic
To find the roots of \( a^2 + 4a - 3 = 0 \):
Using the quadratic formula:
\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{16 + 12}}{2} = \frac{-4 \pm \sqrt{28}}{2} = \frac{-4 \pm 2\sqrt{7}}{2} = -2 \pm \sqrt{7} \]
Thus, the roots are:
\[ a_1 = -2 + \sqrt{7}, \quad a_2 = -2 - \sqrt{7} \]

### Step 6: Determine the Interval
The quadratic opens upwards, so the inequality \( a^2 + 4a - 3 \leq 0 \) holds between the roots:
\[ -2 - \sqrt{7} \leq a \leq -2 + \sqrt{7} \]

### Step 7: Solve the Second Inequality
Now we solve the second inequality:
\[ \sqrt{2} + \sqrt{2 - a} \leq \sqrt{a + 4} \]
Squaring both sides:
\[ (\sqrt{2} + \sqrt{2 - a})^2 \leq a + 4 \]
Following similar steps as before, we will find the range of \( a \) from this inequality.

### Final Solution
After solving both inequalities, we will find the values of \( a \) for which the original equation has solutions.