the number of solution of 2^(cos x)=abs(sin x) in (0,pi) is
If you take the log of both sides of the equation you get
$\displaystyle \ln(2)\cos(x)=\ln(|sin(x)|)$
The sine function from zero to Pi over 2 goes from zero to 1 and is increasing on that interval. The graph is symmetric over the line x=PI/2.
This tells us that
$\displaystyle f(x)=\ln|\sin(x)|\le 0 $ increase from negative infinity to zero at x=Pi/2 and then decreases from 0 to negative infinite at Pi.
the function
$\displaystyle g(x)=\ln(2)\cos(x)$ has one zero at x =Pi/2 and decreases from ln(2) to negative ln(2) on the interval.
So this has one solution when x=Pi/2. Since both functions are monotone decreasing on Pi/2 Pi and
$\displaystyle \lim_{x \to \pi}g(x)=-\ln(2) \quad \lim_{x \to \pi}f(x)=-\infty$
They will cross exactly one more time in the inteval $\displaystyle \left( \frac{\pi}{2},\pi\right)$