Sinx*tgx=1/2

sin (x)*tg (x)=1/2sin (x)*sin (x) /cos (x)=1/2sin^2 (x) /cos (x)=1/2sin^2 (x)=(1/2)*cos (x) 1-cos^2 (x)=(1/2)*cos (x) 2-2cos^2 (x)=cos (x) 2cos^2 (x)+cos (x) -2=0cos (x)=t2t^2+t-2=0D=b^2-4ac=1+16=17t1,2=(-b±sqrt (D) /2at1=(-1-sqrt (17) /4t2=(-1+sqrt (17) /4То есть 1. cos (x)=(-1-sqrt (17) /4<-1 не удовлетворяет ОДЗ 2. cos (x)=(-1+sqrt (17) /4x=±arcos (-1+sqrt (17) /4)+2*pi*nОтвет: x=±arcos (-1+sqrt (17) /4)+2*pi*n