To find the value of \(x\) that maximizes the function \(6 \sin(2x - 25)^\circ\) in the given range \(0^\circ \leq x \leq 180^\circ\), we can look for the critical points of the function within this interval. The maximum will occur at a critical point or at the endpoints of the interval.

First, let's find the critical points by taking the derivative of the function with respect to \(x\):

\(\frac{d}{dx} [6 \sin(2x - 25)^\circ] = 6 \cdot 2 \cos(2x - 25)^\circ \cdot 2 = 24 \cos(2x - 25)^\circ\(

The critical points occur where the derivative is equal to zero:

\(24 \cos(2x - 25)^\circ = 0\(

\(\cos(2x - 25)^\circ = 0\(

To find the solutions for this equation, we look for the angles within the given range \(0^\circ \leq x \leq 180^\circ\) where the cosine function is equal to zero. Cosine is equal to zero at \(90^\circ\) and \(270^\circ\), which corresponds to \(2x - 25 = 90^\circ\) and \(2x - 25 = 270^\circ\):

\(2x - 25 = 90^\circ\(

\(2x = 115^\circ\(

\(x = 57.5^\circ\(

\(2x - 25 = 270^\circ\(

\(2x = 295^\circ\(

\(x = 147.5^\circ\(

Since \(0^\circ \leq x \leq 180^\circ\), the only valid critical point is \(x = 57.5^\circ\).

Now, we need to evaluate the function at the critical point and at the endpoints of the interval:

For \(x = 0^\circ\):

\(6 \sin(2x - 25)^\circ = 6 \sin(-25)^\circ = -1.5\(

For \(x = 57.5^\circ\):

\(6 \sin(2x - 25)^\circ = 6 \sin(2 \cdot 57.5 - 25)^\circ = 6 \sin(110)^\circ = 6 \cdot 1 = 6\(

For \(x = 180^\circ\):

\(6 \sin(2x - 25)^\circ = 6 \sin(335)^\circ = -6 \cdot 1 = -6\(

The maximum value is attained at \(x = 57.5^\circ\), which is \(6\).