2cos2x=корень3sin(3p/2-x)

Question

2cos2x=корень3sin(3p/2-x)

Answer 1

2cos2x=√3sin(3p/2-x)
2cos2x+√3cosx=0
4cos²x+√3cosx-2=0
cosx=a
4a²+√3a-2=0
D=3+32=35
a1=(-√3-√35)/8⇒cosx=(-√3-√35)/8⇒x=π-arccos(√3+√35)/8 +2πn
a2=(-√3+√35)/8⇒cosx=(-√3+√35)/8⇒x=arccos(-√3+√35)/8 +2πn