1
f(x)=sin³(π/2+x)=cos³x
f`(x)=3cos²x*(-sinx)=-3cos²xsinx
f`(π/6)=-3(√3/2)²*1/2=-3*3/4*1/2=-9/8=-1,125
3
cosx/(1+sinx)+sinx/cosx =2
(cos²x+sinx+sin²x)/[cosx(1+sinx)]=2
(1+sinx)/[cosx(1+sinx)]=2
1/cosx=2,sinx≠-1
cosx=1/2
x=+-π/3+2πn,n∈z
-6x=-5π/3
x=-π/3
x=π/3
x=5π/3
Ответ x={-5π/3;-π/3;π/3;5π/3}