Step-by-step Solution:

The given function is: \[ 4 \sin^2 x + 3 \cos^2 x \] We rewrite it using the identity \( \cos^2 x = 1 - \sin^2 x \): \[ 4 \sin^2 x + 3(1 - \sin^2 x) = 4 \sin^2 x + 3 - 3 \sin^2 x \] \[ = \sin^2 x + 3 \] Since \( 0 \leq \sin^2 x \leq 1 \), the maximum value of \( \sin^2 x + 3 \) is: \[ 4 \] Next, we analyze: \[ \sin \frac{x}{2} + \cos \frac{x}{2} \] Using the identity: \[ \sin A + \cos A = \sqrt{2} \sin \left(A + \frac{\pi}{4} \right) \] Since the maximum value of \( \sin \) function is 1, we get: \[ \sin \frac{x}{2} + \cos \frac{x}{2} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} \] From the inequality: \[ - \sqrt{a^2 + b^2} \leq a \sin x + b \cos x \leq \sqrt{a^2 + b^2} \] The maximum value is attained at \( x = \frac{\pi}{2} \). Thus, the given function has a maximum value: \[ 4 + \sqrt{2} \]