Question

2 tgx/1+tg^2x=-1/2 как решить

Answer 1

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+2Пk;  x=-П/12+Пk
2x=-5п/6+2Пk ; x=-5П/12+Пk

Answer 2

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.