2 варинант , надо выполнить

Объяснение:

a)  (23+3х) + (8х-41)=15;

23+3х+8х-41=15;

11х=33;

х=3.

б)  (19+2х) - (5х-11)=25;

19+2х-5х+11=25;

-3х= -5;

x=5/3.

***

в)  (3,2y-1.8)-(5.2y+3.4)=-5.8;

3.2y-1.8-5.2y-3.4=-5.8;

3.2y-5.2y=-5.8+3.4+1.8;

-2y= -0,6

y=0,3.

***

г)  1-(0,5х-15,8)=12,8-0,7х;

1-0,5х+15,8=12,8-0,7х;

-0,5х+0,7х=12,8-15,8-1;

0,2х= -4;

х= -  20.

***

д)  3,8-1,5y+(4.5y-0.8)=2.4y+3;

3.8-1.5y+4.5y-0.8=2.4y+3;

-1.5y+4.5y-2.4y=3-3.8+0.8;

0.6y =0;

y=0/0.6;

y=0.

***

е)  4.2y+0.8=6.2y-(1.1y+0.8)+1.2;

4.2y+0.8=6.2y-1.1y-0.8+1.2;

4.2y-6.2y+1.1y=-0.8+1.2-0.8;

-0.9y=-0.4;

y=4/9.

1)    ;
sin2x - (1-sin²x)  =0 ;
2sinxcosx -cos²x =0 ;
cosx(2sinx -cosx) =0 ;
[cosx =0 ;2sinx-cosx =0.⇔ [cosx =0 ;sinx=(1/2)cosx.⇔[cosx =0 ;tqx=1/2.
[ x=π/2 +πn ; x =arctq1/2+πn , n∈Z.

2)   ;
ctq2x*cos²x - ctq2x*sin²x =0 ;
ctq2x*(cos²x - sin²x) =0 ;
ctq2x*cos2x =0 ;
sin2x =0  * * *cos2x = ± 1 ≠0→ ОДЗ * * * 
2x =πn , n∈Z ;
x =(π/2)*n , n∈Z .

3)   ;
3sin²x/2 -2sinx/2 =0 ;
3sinx/2 (sinx/2 -2/3) =0 ;
[sinx/2 =0 ; sinx/2 =2/3 .⇒[x/2 =πn ; x/2= arcsin(2/3) +πn ,n∈Z.⇔
[x =2πn ; x= 2arcsin(2/3) +2πn ,n∈Z.

4)  ;
* *cos2α =cos²α -sin²α =cos²α -(1-sin²α)=2cos²α -1⇒1+cos2α=2cos²α * *
cos3x = 1+cos2*(3x) ;  * * * α = 3x  * * *
cos3x = 2cos²3x ; 
2cos²3x -cos3x =0 ;
2cos3x(cos3x -1/2) =0 ;
[cos3x =0 ; cos3x =1/2 ⇒[3x=π/2+πn ; 3x= ±π/3+2πn ,n∈Z.⇔
[x=π/6+πn/3 ; x= ±π/9+(2π/3)*n ,n∈Z.