Question

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

Answer 1

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
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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.

Answer 2

arccos(1+√3)/2√2 можно упростить