Question

y(z)= 3 cos^2 x + 3√2 sin x+ 4
 y`(z)=0

Answer 1

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