Question

Найти корни, принадлежащие промежутку [-2π; 2π)

Answer 1

1)cos(4x/3-3π/4)=√2/2
4x/3-3π/4=±π/4+2πk
4x/3=3π/4±π/4+2πk ;•3/4
x=9π/16±3π/16+3πk/2

x1=9π/16+3π/16+3πk/2=11π/16+3πk/2
x2=9π/16-3π/16+3πk/2=3π/8+3πk/2

2)-2π≤11π/16+3πk/2<2π
-32π≤11π+24πk<32π
-43π≤24πk<21π
-43/24≤k<21/24
-43/24≤k<7/8
-1 19/24≤k<7/8
k={-1;0}

2)-2π≤3π/8+3πk/2<2π
-16π≤3π+12πk<16π
-19π≤12πk<13π
-19/12≤k<13/12
k={-1;0;1}