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

Answer 2

It seems like you have provided two equations:

1. Y(z) = 3cos^2(x) + 3√2sin(x) + 4
2. y'(z) = 0

The first equation represents a function Y(z) in terms of x, where cos^2(x) represents the square of the cosine of x, and sin(x) represents the sine of x. The equation also includes constants (√2 and 4).

The second equation represents the derivative of a function y(z) with respect to z, and it is set equal to zero, indicating that the function y(z) is constant with respect to z.

However, it's worth noting that there is no direct connection between the two equations you provided. If you have any specific question or if you would like further assistance with these equations, please let me know.