Question

Найдите критические точки функции: y=x-2 cosx

Answer 1

Ответ:

x = (-1)^k × π/6 + πk , k € Z

Объяснение:

y = x-2cosx

y ` = (x-2×cosx) ` = (x) ` - (2×cosx) ` = (x) ` - (2) ` × (cosx) + (2) × (cosx) ` = 1 - 2sinx

y ` = 0

1 - 2sinx = 0

-2sinx = -1

sinx = 1/2

x = (-1)^k × arcsin(1/2) + πk , k€ Z

x = (-1)^k × π/6 + πk , k € Z