- 2sin²x - √3sin2x =0;- 2sin²x - √3*2sinx *cosx=0;-2sinx(sinx -√3cosx) = 0 ;[ sinx = 0 ; sinx -√3cosx =0 ;a) sinx = 0 ⇒ x = π*k , k∈Z ;.b) sinx -√3cosx =0 ⇔tqx =√3 ⇒x =π/3 + π*k , k∈Z.ответ : π*k ; π/3 + π*k , k∈Z.********************************************2sin²x = 1 +cosx ; - (1 -2sin²x) = cosx ;- cos2x = cosx ; cos2x +cosx =0 ; * * * * * cosα +cosβ =2cos(α+β)/2* cos(α - β)/2 * * * * *2cos3x/2*cosx/2 =0 ;cos3x/2 = 0 ⇒3x/2 =π/2+π*k , k∈Z⇔x = π/3+2π/3*k , k∈Z ;cosx/2 =0⇒x/2 =π/2+π*k , k∈Z⇔x = π+2π*k ,k∈Z.ответ : π/3+2π/3*k ; π+2π*k , k∈Z.