Consider the equation.

The solutions of this equation over the interval [0, π] are 2pi/3, 5pi/6 pi/6, 2pi/3 7pi/6, 11pi/6 pi/6, 5pi/6

Show that (x^3) / 2=sin [ πx/4 + π / 2) has a solution over the interval [ 0,2]. Oct 20 | Fresh from Anchorage, AK ... Mark M. Carson, CA · Solve the given equation over the interval [ 0, 2π) : 2 cos x - sin2 x = cos2 x ... A. x = (5pi/6) and x = (7pi/6) B. x = (2pi/3) and x = (4pi/3) C. x = (pi/6) and x = (11pi/6) D. x = (pi/3)