Question

Прошу помочь решить задачу №4

Answer 1

f(x)=2cos2x+3sinx;cosπ=-1;sinπ/2=1
x=π/2
f(π/2)=2cos(2*π/2)+3sin(π/2)=2•cosπ+3
•sinπ/2=-2+3=1

Answer 2

f(x)=2 cos2*p/2 +3sin p/2 = 2cosp +3sin p/2 = 2*(-1)+3*1= -2+3=1

B)