2cos²x+sin2x=sin(x-3π/2)-cos(π/2+x)
2cos²x+sin2x=-sin(3π/2-x)-cos(π/2+x) =cosx+sinx
2cos²x+2sinxcosx-(cosx+sinx)=0
2cosx(cosx+sinx)-(cosx+sinx)=
(cosx+sinx)·(2cosx-1)=0
(cosx+sinx)=0? 2cosx=1
1+tgx=0 cosx=0,5
tg x= -1 x=(+-)π/3+2πn,n∈Z
x=-π/4+πn,n∈Z
Ответ:
x=-π/4+πn,n∈Z,
x=(+-)π/3+2πn,n∈Z