cos2x=sin(3pi/2-x) решить уравнение

cos2x = sin(3pi/2 - x)cos2x = - cosx cos2x + cosx =  02cos^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 = 1/2t2 = ( - 1 - 3)/4 = - 4/4 = - 1cosx = 1/2x = ± arccos(1/2) + 2pikx = ± pi/3 + 2pik.k ∈ Zcosx = - 1x = pi + 2pik.k ∈ Z Ответ:x = ± pi/3 + 2pik.k ∈ Zx = pi + 2pik.k ∈ Z