Решите, cosx-2sinx=0 sin3x-sin7x=0 7sin2x-2sinx=0 sin2x-2sinx=0 5cos2x-14cos2x+8 3cos2x-14cosx+8

cosx-2sinx=0;
tqx =1/2 ;
x =arctq1/2+π*k ,k∈Z.

sin3x-sin7x=0 ;
-2sin2x*cos5x=0;
[sin2x =0  ;cos5x =0.
[x =π/2*k,  x =π/10 +π/5*k ; k∈Z.

7sin2x-2sinx=0 ;
14sinxcosx -2sinx =0 ;
14sinx(cosx -1/7) =0 ;
[sinx=0 ; cosx =1/7.[  x =π*k ; x =(+/-) arccos1/7 +2π*k  ,k∈Z
  или    ?  
7sin²x -2sinx =0;
7sinx(sinx -2/7) =0 ;
[sinx=0 ; sinx =2/7.[ x =π*k ; x =(-1)^k* arcsin2/7 +π*k  ,k∈Z.

sin2x-2sinx=0 ;
2sinxcosx -2sinx=0;
2sinx(cosx -1) = 0;
[sinx =0 ;cosx =1. [x= x = π*k ;x =2π*k , k∈Z.
  или   ? 
sin²x -2sinx =0;
sinx(sinx-2) =0⇔ sinx = 0⇒ x = π*k, k∈Z .

5cos²x-14cos2x+8 = 0;
[cosx =4/5 ;cosx =2. ⇔cosx=4/5⇒x=(+/-)arccos(4/5) +2π*k ,k∈Z.

3cos²x-14cosx+8 =0 ;
[cosx =2/3 ; cosx =4 ⇔cosx=2/3⇒x=(+/-)arccos(2/3) +2π*k ,k∈Z.