Question

rozwiąż równania:

a) (1-sin2x)/2=1

b) sin(2x-pi/3)=1

Answer 1

(1-sin2x)/2=1

1-sin2x=2

sin2x=-1

sin2x=sin(-90)

2x=-90+k·360

x=-45+k·180 k∈C

 

sin(2x-pi/3)=1

sin(2x-pi/3)=sin(π/2)

2x-pi/3=π/2+2kπ

2x=5π/6+2kπ

x=5π/16+kπ k∈C