Question

2cos²x=√3sin(3П/2+x)

Answer 1

2cos²x=√3sin(3π/2+x)
2cos²x= -√3cosx
2cos²x+√3cosx=0
cosx(2cosx+√3)=0
cosx=0                         или                2cosx+√3=0
x=π/2+πn, n∈Z                                 cosx=√3/2
                                                         x=+-π/6+2πn, n∈Z

Answer 2

Спасибо тебе огромное))))