Answer
An absolute maximum at $\pi/2$, and an absolute minimum at $x=3\pi/2$.
Th. 1 does not apply, so the answer is not inconsistent with it.
Work Step by Step
See image for the graph of $y=f(x)$.
The point ($\pi/2,3$) is on the graph and its y-coordinate is greater than any other on the graph.
So, $y=f(x)$ has an absolute maximum at $x=\pi/2$.
The point ($3\pi/2,-3$) is on the graph and its y-coordinate is lower than any other on the graph.
So, $y=f(x)$ has an absolute minimum at $x=3\pi/2$.
Theorem 1 demands that "$f$ is continuous on a closed interval $[a, b]$,...", so it does not apply here (the domain here is an open interval). Thus, there are no inconsistencies.
