Question

найдите критические точки функции у=2cosx-x

Answer 1

y' = (2cos(x) -x)' = 2(cos(x))' -(x)' = -2sin(x) -1

-2sin(x) -1 = 0

sin(x) = -1/2

x = (-1)ⁿ arcsin(-1/2) +πn, n ∈ Z

x = (-1)ⁿ⁺¹ arcsin(1/2) +πn, n ∈ Z

x = (-1)ⁿ⁺¹ π/6 +πn, n ∈ Z