Cos2x=sin(3pi/2-x) решить уравнение

cos2x = sin(3pi/2 - x) cos2x = - cosx  cos2x + cosx =  0 2cos^2x + cosx - 1 = 0  пусть cosx = t, t ∈ [ -1; 1]  2t^2 + t - 1 = 0  D = 1 + 4*2 = 9  t1 = ( - 1 +3)/4 = 2/4 = 1/2 t2 = ( - 1 - 3)/4 = - 4/4 = - 1 cosx = 1/2 x = ± arccos(1/2) + 2pik x = ± pi/3 + 2pik.k ∈ Z cosx = - 1 x = pi + 2pik.k ∈ Z  Ответ: x = ± pi/3 + 2pik.k ∈ Z x = pi + 2pik.k ∈ Z