we need to sum the roots of the equation sin2x=92 between x=0 and x=2π. These roots, all of which must be roots of the given equation, are
x=2arcsin92,2π−arcsin92,22π+arcsin92, and 23π−arcsin92
and their sum is 3π.
OR
For any b>2 the solutions of y2−by+1=0 are y1,y2=2b±b2−4, which are distinct and positive. These solutions are reciprocals because their product must be 1. Since the solutions yi=tanxi are reciprocals,
tanx2=y2=y11=tanx11=cotx1=tan(2π−x1)
Thus, y1 and y2 are tangents of two distinct, complementary, first-quadrant angles, x1 and x2=2π−x1. Since tan(x+π)=tanx, there are four values of x between 0 and 2π:x1,π+x1,2π−x1 and 23π−x1. Their sum is 3π.