If you take the log of both sides of the equation you get

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

increase from negative infinity to zero at x=Pi/2 and then decreases from 0 to negative infinite at Pi.

the function

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

They will cross exactly one more time in the inteval