РЕШИТЕ Безотлагательно, ПОЖАЛУЙСТА!!!!1)sin2x=sin(x-pi/3)2)cos(x-pi/6)=cos(pi/5)3)cos2x=sin(pi/3+x)

task/29916604/29916224

1.   sin2x = sin(x -/3) sin2x + sin(/3 -x) 2sin(x/2 +/6)*cos(3x/2 -/6) =0

[ sin(x/2 +/6) =0 ; cos(3x/2 -/6) =0 . [ x/2 +/6 =n ; 3x/2 -/6 =/2 + n , n .

[x= - /3 + 2n ;  x =4/9 + (2/3)*n , n .

2. cos(x - /6) = cos(/5) cos(x - /6) - cos(/5) =0

-2sin( (x-/6-/5)/2 )*sin( (x-/6+ /5)/2) =0 sin( (x-11/30) /2)*sin((x+/30)/2)=0

[ sin( (x-11/30) /2) =0 ; sin((x+/30)/2)=0.[ (x-11/30)/2 =n ;  (x+/30)/2=n , n

[x = 11/30 +2n ;  x =  - /30 +2n , n .

3.  cos2x = sin(/3 +x) cos2x = cos(/2 -(/3 +x) ) cos2x - cos(/6 -x)  =0

-2sin( (3x -/6) /2) *sin( ( x +/6) /2) =0 [sin( (3x -/6) /2) =0 ;sin( ( x +/6) /2)=0.

[ ( 3x -/6)/2 =n ; (x +/6)/2 =n, n

[ x=/18+(2/3)*n ; x = - /3 +2n ,n .

* P.S. sin+sin=2sin((+)/2)*cos((- )/2) ;cos-cos =-2sin(( -)/2)*sin((+)/2) ; sin =cos(/2 - ) *