Решение
4x+3cos4x=0
cos4x(cos4x + 3) = 0
1) cos4x = 0
4x = π/2 + πk, k∈Z
x = π/8 + πk/4, k∈Z
2) cos4x + 3 = 0
cos4x = - 3 не удовлетворят условию I cosxI ≤ 1
Ответ: x = π/8 + πk/4, k∈Z
14x+5cosx-1=0
14(1 - cos²x) + 5cosx - 1 = 0
14 - 14cos²x + 5cosx - 1 = 0
14cos²x - 5cosx - 13 = 0
I cosx I ≤ 1
cosx = t
14t² - 5t - 13 = 0
D = 25 + 4*14*13 = 753
t₁ = (5 - √753)/28
t₂ = (5 + √753)/28 не удовлетворят условию I cosxI ≤ 1
cosx = (5 - √753)/28
x = (+ -)arccos(5 - √753)/28 + 2πn, n∈Z