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