Sin² x + sin²2x + sin²3x =1,5
sin^2α =(1-cos2 α )/2
(1-cos2x)/2 + (1-cos4x)/2 +(1-cos6x)/2= 3/2;
cos2x+cos4x+cos6x=0
(cos2x+cos6x)+cos4x=0
Формула
cos α +cos β =2cos( α + β )/2 * cos( α - β )/2
2cos4x *cos2x+cos4x=0
cos4x*(2cos2x+1)=0
cos4x=0 ⇒ 4x=(π/2)+πk, k ∈ Z ⇒ [b] x=(π/8)+(π/4)k, k ∈ Z[/b]
или
2сos2x+1=0 ⇒ cos2x=-1/2 ⇒ 2x= ± (2π/3)+2πn, n ∈ Z ⇒
[b]x=± (π/3)+πn, n ∈ Z[/b]