text: math help trig..? find all the solutions of the equation in the interval (0,2π) algebraically. cos^3 x=cosx  and 2sin^2 x +sinx+1=0 those are just disguised factoring problems.  consider:  cos^3x = cos x  cos^3 x - cos x = 0  cos x(cos^2 x - 1) = 0  -cos x(sin^2 x) = 0  so cos x = 0 or sin^2 x = 0  and solve from there.  for the other, note that the equation is quadratic in sin x.  try the substitution y = sin x:  2y^2 + y + 1 = 0  solve that for y and resubstitute sin x.