Question

5cos2x-3=0

Напишите решение

Answer 1

5cos2x-3=0

cos2x=3/5

2x=±arccos3/5+2kπ; k∈Z;

x=±(arccos3/5)/2+kπ; k∈Z; 

Answer 2

5cos(2x)-3=0

5cos(2x)=3

cos(2x)=3/5

2x=±arccos(3/5)+2*pi*n

x=±(1/2)*arccos(3/5)+pi*n