Use cos2x+sin2x=1. The equation becomes sin3x+1−sin2x=0. Write s for sinx. The cubic s3−s2+1 is −1 at s=−1, is 1 at s=0, and is 1 at s=1. The turning points are at s=0 and s=2∕3, and the value at 2∕3 is 23∕27>0. So this cubic has one (negative) root.
There are two solutions for sinx=s in the given range.