Question

найти нули и область значения функции f(x)=cos (9x/11) - 1/2

 

Answer 1

cos(9x/11)-1/2=0
cos(9x/11)=1/2
9x/11 = +-arccos(1/2) +2пk
9x/11 = +-п/3 + 2пk
9x = +-11п/3 + 22пk
x = +-11п/27 + 22пk/9 - нули функции

-1<=cos(9x/11)<=1
-1.5<=cos(9x/11)-1/2<=0.5

Ответ: x=+-11п/27 + 22пk/9 ; [-1.5;0.5]