Y (z) = 3 cos^2 x + 3√2 sin x + 4

y' (z) = 0

Решение

Y (x) = 3 cos^2 x + 3√2 sin x + 4

y' = 6cosx * (- sinx) + 3 √2cosx = - 6sinxcosx + 3√2sinx

Если y' (x) = 0, то

- 6sinxcosx + 3√2sinx = 0

6sinxcosx - 3√2sinx = 0

3sinx (2cosx - √2) = 0

1) 3 sinx = 0

sinx = 0

x = πk, k ∈ Z

2) 2cosx - √2 = 0

cosx = √2/2

x = + - arccos (√2/2) + 2πn, n ∈ z

x = + - (π/4) + 2πn, n ∈ z