А) при x=2
Б) при х=-2, -1, 3
(sinx+sin3x)-(sin2x+sin4x)=0
2sin2xcosx-2sin3xcosx=0
2cosx*(sin2x-sin3x)=0
cosx=0⇒x=π/2+πn,n∈z
sin2x-sin3x=0
-2sin(x/2)cos(5x/2)=0
sin(x/2)=0⇒x/2=πk⇒x=2πk,k∈z
cos(5x/2)=0⇒5x/2=π/2+πm⇒x=π/5+2πm/5,m∈z
S₁₉=19(a₁+a₁₉)/2
19(a₁+a₁₉)/2=171
19(a₁+a₁₉)=342
a₁+a₁₉=342/19=18
a₁₀=(a₁+a₁₉)/2=18/2=9
отв:9