Answer
$ c\approx -0.5756$
Work Step by Step
Rolle's Theorem
Let $f$ be continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$ .
If $f(a)=f(b)$ , then there is at least one number $c$ in $(a, b)$ such that $f^{\prime}(c)=0.$
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Check the conditions for the theorem:
$f$ is continuous on $[-1,\ 0].$
$f$ is differentiable on $(-1,0)$.
$f(-1)=f(0)=0$
Rolle's Theorem applies, a value x=c exists such that $f^{\prime}(c)=0$
$ f^{\prime}(x)=\displaystyle \frac{1}{2}-(\cos\frac{\pi x}{6})\cdot\frac{\pi}{6}\quad$ (chain rule on the second term)
$f^{\prime}(c)=0$
$\displaystyle \frac{1}{2}-\frac{\pi}{6}\cos\frac{\pi c}{6}=0$
$\displaystyle \cos\frac{\pi c}{6}=\frac{3}{\pi}$
$\displaystyle \frac{\pi c}{6}=\arccos\frac{3}{\pi}$
... The RHS wil return a positive value, but we need to see if there is a negative, in the interval $(-1,0)$, so we change the sign ...
$c=-\displaystyle \frac{6}{\pi}\arccos\frac{3}{\pi}\approx-0.5756$ rad
$ c\approx -0.5756$
