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<em>2ctgα(1-cos²α)=2ctgα*sin²α=2cosα*sinα=sin2α.</em>
решение задания смотри на фотографии
6х-3+12=х;6х-х=-12+3;5х=-9;х=1,8
(1/3*1/2m+1/3*(-3)-m*1/2m-m*(-3)
(1/3*1/2m+1*(-1)-m*1/2m-m*(-3)
(m/2*3+1*(-1)-m*1/2m-m*(-3)
1/6m+1*(-a)-m=1/2m-m*(-3)
(1/6m-1-m*1/2m-m*(-3)
(1/6m-1-1/2m^2-m*(-3)
(1/6m-1-1/2m^2+3m)
(-1/2m^2+1/6m+3m-1)
(-1/2m^2+1/6m+3m*6/6-1)
(-1/2m^2+1/6m+18m/6-1)
(-1/2m^2+1/6(m+18m)-1)
(-1/2m^2+19m/6-1)
-1/2m^2+19m/6-1 ответ!