2 tgx/1+tg^2x=-1/2 как решить

tgx=t2t/(1+t^2)=-1/24t=-1-t^2t^2+4t+1=0t=-2+-sqrt(3)t=sqrt(3)-2t=-sqrt(3)-2x=arctg(√3-2)+Пкх=-arctg(√3+2)+Пк(2sinx/cosx)/(1+sin^2x/cos^2x)=-1/2(2sinx/cosx)/1/cos^2x=-1/22sinxcosx=-1/2sin2x=-1/22x=-П/6+2Пk;  x=-П/12+Пk2x=-5п/6+2Пk ; x=-5П/12+Пk

2*tgx/(1+tg²x)=-1/2

2*tgx/(1+(sin²x/cos²x))=-1/2

2*tgx/((cos²x+sin²x)/cosx)=-1/2

2*tgx/(1/cos²x)=-1/2

(2*sinx/cosx)*cos²x=-1/2

2*sinxcosx=-1/2

sin(2x)=-1/2

2x=7π/6+2πn |÷2 x₁=7π/12+πn

2x=11π/6+2πn |÷2 x₂=11π/12+πn

Ответ: x₁=7π/12+πn x₂=11π/12+πn.