the least value of 'a' for which the equation `2sqrt(a) sin^(2) x+sqrt(a-3) sin 2x=5+sqrt(a)` has at least one solution is _______.

Explanation

To solve the equation \( 2\sqrt{a} \sin^2 x + \sqrt{a - 3} \sin 2x = 5 + \sqrt{a} \) and find the least value of \( a \) for which it has at least one solution, we can follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ 2\sqrt{a} \sin^2 x + \sqrt{a - 3} \sin 2x = 5 + \sqrt{a} \]
Recall that \( \sin 2x = 2 \sin x \cos x \) and \( \sin^2 x = 1 - \cos^2 x \). We can express \( \sin 2x \) in terms of \( \sin x \) and \( \cos x \).

### Step 2: Substitute \( \sin 2x \)
Substituting \( \sin 2x \) gives us:
\[ 2\sqrt{a} \sin^2 x + \sqrt{a - 3} (2 \sin x \cos x) = 5 + \sqrt{a} \]
This simplifies to:
\[ 2\sqrt{a} \sin^2 x + 2\sqrt{a - 3} \sin x \cos x = 5 + \sqrt{a} \]

### Step 3: Rearranging the equation
Rearranging the equation, we have:
\[ 2\sqrt{a - 3} \sin x \cos x - \sqrt{a} (1 - 2\sin^2 x) = 5 \]
Using the identity \( 1 - 2\sin^2 x = \cos 2x \), we can rewrite it as:
\[ \sqrt{a - 3} \sin 2x - \sqrt{a} \cos 2x = 5 \]

### Step 4: Analyze the equation
The equation can be expressed in the form:
\[ R \sin(2x + \phi) = 5 \]
where \( R = \sqrt{(\sqrt{a - 3})^2 + (\sqrt{a})^2} \) and \( \tan \phi = \frac{\sqrt{a}}{\sqrt{a - 3}} \).

### Step 5: Find the condition for at least one solution
For the equation \( R \sin(2x + \phi) = 5 \) to have at least one solution, we need:
\[ R \geq 5 \]
Calculating \( R \):
\[ R = \sqrt{(a - 3) + a} = \sqrt{2a - 3} \]
Thus, the condition becomes:
\[ \sqrt{2a - 3} \geq 5 \]

### Step 6: Solve the inequality
Squaring both sides gives:
\[ 2a - 3 \geq 25 \]
\[ 2a \geq 28 \]
\[ a \geq 14 \]

### Conclusion
The least value of \( a \) for which the equation has at least one solution is:
\[ \boxed{14} \]