You need to evaluate the derivative of the given function, such that:

`f'(x) = (x/2 − sin(x*pi/6))' => f'(x) = 1/2 - pi/6*cos(x*pi/6)`

You need to find the value of `c in [-1,0]` such that `f'(c) = 0` , hence, substituting c for x in equation `f'(x) = 0` yields:

`1/2 - pi/6*cos(c*pi/6) = 0 =>- pi/6*cos(c*pi/6) = -1/2`

`pi/3*cos(c*pi/6) = 1 => cos(c*pi/6) = 1/(pi/3)`

`cos(c*pi/6) = 3/pi => cos(c*pi/6) = 0.9549 => c*pi/6 = +-cos^(-1)(0.9549)`

`c*pi/6 = +-0.3014 => c = +-6*0.3014/pi => c = +-0.5756`

**Since the value `c = 0.5756` is not in interval `[-1,0], ` hence, the only valid solution to the equation `f'(c) = 0` is `c = -0.5756` .**

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