9cos2x+20cosx+7=0 Решите плиз надо очень!!

9(2cos^2x - 1) + 20cosx + 7 = 0  18cos^2x - 9 + 20cosx + 7 = 0  18cos^2x + 20cosx - 2 = 0  /:2 9cos^2x + 10cosx - 1 = 0  cosx = t 9t^2 + 10t - 1 = 0  D = 100 - 36 = 64 t1 = ( - 10 + 8)/18 = - 2/18 = - 1/9 t2 = ( - 10 - 8)/18 = - 18/18 = - 1 1) cosx = - 1/9 x = ± arccos(-1/9)+ 2pik x = ± (pi - arccos1/9) + 2pik, k ∈Z 2) cosx = - 1 x = pi + 2pik, k ∈Z