Question

cos2x+корень 32=0 помогите пжл,прошу:)

Answer 1

cos2x+√3/2=0
cos2x= -√3/2
2x=+-arccos(-√3/2)+2πk
2x=+-π-π/6+2πk
2x=+-5π/6+2πk
x=+-5π/12+πk

Answer 2

cos2x+корень 32=0

cos2x=-√(3/2)
2x=arccos(√(3/2)+2πk
x=1/2*(arccos(√(3/2)+2πk)