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Verified answer

1) 
sin2x = 3sinx*cos²x ;
sin2x =(3/2)*2sinx*cosx *cosx ;  * * * 2sinx*cosx =sin2x * * *
sin2x =(3/2)*sin2x*cosx  ;
2sin2x = 3sin2x*cosx ; 
2sin2x -3sin2x*cosx = 0 ;
3sin2x( 2/3 - cosx) =0 ;
[ sin2x =0 ;           [ 2x =πk , k ∈ Z 
[ 2/3 -cosx =0 .    [ cosx = 2/3.

[ x =(π/2)*k , k ∈ Z  ;
[ x = ± arccos(2/3) +2πn , n∈ Z .
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2)
sin4x= sin2x  * * * sin4x =sin2*2x = 2sin2x*cos2x * * *
 2sin2x*cos2x =sin2x ;
 2sin2x*cos2x - sin2x =0  ;
2sin2x(cos2x - 1/2) =0  ;
[ sin2x = 0 ;               [ 2x  =πk , k ∈ Z ;
[cos2x =1/2 .            [ 2x =±π/3 +2πn , n∈Z.

[ x  =(π/2)*k , k ∈ Z ;
[ x =±π/6 +πn , n∈Z.
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3) 
cos2x+cos²x =0 ;  
* * * cos2x =cos²x - sin²x =cos²x -(1 -cos²x) =2cos²x -1⇒cos²x =(1+cos2x)/2 * * 
cos2x+(1+cos2x)/2 =0 ; 
2cos2x +1 +cos2x =0 ; 
3cos2x = -1 ;
cos2x = -(1/3) ; 
2x =±(π -arccos(1/3)) +2πk , k ∈Z.
x =±0,5(π -arccos(1/3)) +πk , k ∈Z.
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4)
sin2x =cos²x ; 
2sinx*cosx = cos²x ;
2sinx*cosx - cos²x =0  ;
cosx(2sinx -cosx) =0 ;
[ cosx =0 ;                  [ x =π/2 +πk , k∈Z;     [x=π/2 +πk , k∈Z;
[2sinx - cosx =0 .       [ 2sinx = cosx .           [ tgx = 1/2 .

[x=π/2 +πk , k∈Z;
[x = arctg(1/2) +πn , n∈Z . 

sedinalana 1
sin2x=3sinxcos²x
2sin2x-1,5sin2xcosx=0
sin2x*(1-1,5cosx)=0
sin2x=0⇒2x=πk,k∈z⇒x=πk/2,k∈z
1-1,5cosx=0
1,5cosx=1
cosx=2/3⇒x=+-arccos2/3+2πk,k∈z
2
sin4x=sin2x
sin4x-sin2x=0
2sinx*cos3x=0
sinx=0⇒x=πk,k∈z
cos3x=0⇒3x=π/2+πk,k∈z⇒x=π/6+πk/3,k∈z
3
cos2x+cos²x=0
2cos²x-1+cos²x=0
cos²x-1=0
(cosx-1)9cosx+1)=0
cosx=1⇒x=2πk,k∈z
cosx=-1⇒x=3π/2+2πk,k∈z
4
sin2x=cos²z
2sinxcosx-cos²x=0
cosx*(2sinx-cosx)=0
cosx=0⇒x=π/2+πk,k∈z
2sinx-cosx=0/cosx
2tgx-1=0
2tgx=1
tgx=1/2
x=arctg1/2+πk,k∈z