Сколько корней, принадлежащих [0;П] , имеет уравнение cos6x+cos4x=0

cosa + cosb = 2 * cos (a - b)/2 * cos (a + b)/2 cos6x + cos4x=0 2cosx*cos5x =  0 cosx = 0 ==> x = pi/2 + pik cos5x = 0 ===> x = pi/10 + pik/5 отбор 1) 0 ≤ pi/2 + pik ≤ pi 0 ≤ 1/2 + k ≤ 1 - 1/2 ≤ k ≤ 1/2 k = 0 ==> x = pi/2 2) 0 ≤ pi/10 + pik/5 ≤ pi 0 ≤ 1/10 + k/5 ≤ 1 - 1/10 ≤ k/5 ≤ 9/10 - 1/2 ≤ k ≤ 9/2  = 4,5 k = 0; 1; 2; 3; 4 x1 = pi/10 x2 = pi/10 + pi/5 = 3pi/10 x3 = pi/10 + 2pi/5 = pi/2 x4 = pi/10 + 3pi/5 = 7pi/10 x5 = pi/10 + 4pi/5 = 9pi/10 ОТВЕТ: pi/2 + pik, k ∈ Z pi/10 + pik/5, k ∈ Z