2 tgx/1+tg^2x=-1/2 как решить

Tgx=t

2t / (1+t^2) = - 1/2

4t=-1-t^2

t^2+4t+1=0

t=-2+-sqrt (3)

t=sqrt (3) - 2

t=-sqrt (3) - 2

x=arctg (√3-2) + Пк

х=-arctg (√3+2) + Пк

(2sinx/cosx) / (1+sin^2x/cos^2x) = - 1/2

(2sinx/cosx) / 1/cos^2x=-1/2

2sinxcosx=-1/2

sin2x=-1/2

2x=-П/6+2 Пk; x=-П/12+Пk

2x=-5 п/6+2 Пk; x=-5 П/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.