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

РешениеY(x)= 3 cos^2 x + 3√2 sin x+ 4y` = 6cosx * (- sinx) + 3√2cosx = - 6sinxcosx + 3√2sinxЕсли y`(x)=0, то - 6sinxcosx + 3√2sinx = 0 6sinxcosx - 3√2sinx = 03sinx(2cosx - √2) = 01) 3 sinx = 0sinx = 0x = πk, k ∈ Z2)  2cosx - √2 = 0cosx = √2/2x = +-arccos(√2/2) + 2πn, n ∈ zx = +-(π/4) + 2πn, n ∈ z