ПОМОГИТЕ ПОЖАЛУЙСТА!решите уравнения
1)sin8x-cos6x=корень из 3*(sin6x+cos8x)
2)cos2x=((1+корень из 3)/2)*(cosx+sinx)

arccos(1+√3)/2√2 можно упростить

1) sin8x -√3cos8x  = √3sin6x+cos6x ;2sin(8x -π/3) =2sin(6x+π/6) ;[ (8x -π/3)-(6x+π/6) =2πn ; (8x -π/3)+(6x+π/6)  =π+2πn , n∈Z[x=π/4 + πn ;x =π/12+ πn/7 , n∈Z--------------------------2)   cos2x =((1+√3)/2) *(cosx+sinx) ; cos²x - sin²x =((1+√3)/2) *(cosx+sinx) ;(cosx-sinx)(cosx+sinx) - ((1+√3)/2) *(cosx+sinx) =0 ;(cosx+sinx)( cosx-sinx -(1+√3)/2) =0 ;[ cosx+sinx =0 ;cosx-sinx =(1+√3)/2 .[ tqx = -1 ;√2 * cos(x+π/4) =(1+√3)/2 .[x = -π/4 +πn ; x +π/4 =±arccos(1+√3)/2√2 +2πn , n∈Z.[x = -π/4 +πn ; x = -π/4±arccos(1+√3)/2√2 +2πn , n∈Z.