Решите уравнение а)cosx=-1 б)2sin^2x-sinx-1=0 в) cos^2x-V3sinxcosx=0

a) cosx = -1 x = π + 2πk, k€Z б) 2sin²x - sinx - 1 = 0 Пусть t = sinx, t€[-1; 1]. 2t² - t - 1 = 0 D = 1 + 2•4 = 9 = 3² t1 =(1 + 3)/4 = 1 t2 = (1 - 3)/4 = -1/2 Обратная замена: sinx = 1 x = π/2 + 2πk, k€Z sinx = -1/2 x = (-1)ⁿ+¹π/6 + πk, k€Z. в) cos²x - √3sinxcosx = 0 cosx(cosx - √3sinx) = 0 cosx = 0 x = π/2 + πk, k€Z cosx = √3sinx 1 = √3tgx tgx = √3/3 x = π/6 + πk, k€Z. Ответ: x = π/2 + πk, k€Z; x = π/6 πk, k€Z.