Since we can apply Rolle's Theorem:
[tex]\begin{gathered} f^{\prime}(x)=-\sin (x) \\ so\colon \\ f^{\prime}(x)=0 \\ -\sin (x)=0 \end{gathered}[/tex]
Take the inverse sine of both sides:
[tex]\begin{gathered} x=\sin ^{-1}(0) \\ x=\pi n \\ n\in\Z \end{gathered}[/tex]
Since it is for the interval:
[tex]\lbrack\pi,3\pi\rbrack[/tex]
The solutions are:
[tex]x=\frac{3\pi}{2},\frac{5\pi}{2}[/tex]
Answer:
[tex]\begin{gathered} c=\frac{3\pi}{2},\frac{5\pi}{2} \\ or \\ c\approx4.71,7.85 \end{gathered}[/tex]
