A

To solve for the values of \( \sin \theta \) and \( \cos \theta \) given that \( \tan \theta = \sqrt{3} \), we start by recalling the relationship between tangent, sine, and cosine: \( \tan \theta = \frac{\sin \theta}{\cos \theta} \).
Since \( \tan \theta = \sqrt{3} \), we can represent this as \( \frac{\sin \theta}{\cos \theta} = \sqrt{3} \).
Let's select a right triangle where the opposite side is \( \sqrt{3} \) and the adjacent side is \( 1 \). Using the Pythagorean theorem:
\[ \text{hypotenuse} = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2. \]
Thus, we can find \( \sin \theta \) and \( \cos \theta \):
\[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\sqrt{3}}{2}, \quad \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{2}. \]
Now substitute these values into the product \( 2 \sin \theta + 3 \cos \theta \):
\[ 2 \sin \theta + 3 \cos \theta = 2 \left(\frac{\sqrt{3}}{2}\right) + 3 \left(\frac{1}{2}\right) = \sqrt{3} + \frac{3}{2}. \]
So, the value of \( 2 \sin \theta + 3 \cos \theta \) evaluates to a positive number, and the condition given that it's negative cannot be satisfied under these conditions, meaning no real solution exists if we interpret \( \theta \) in the standard range. Therefore, we state the output based solely on given conditions and interpretations.