(sinx-1)/(1+cos2x)=(sinx-1)/(1-cosx)
(sinx-1)/2cos²x-(sinx-1)/(1-cosx)=0
{cosx≠0⇒x≠π/2+πk,k∈z
{1-cosx≠0⇒cosx≠1⇒x≠2πk,k∈z
(sinx-1)(1-cosx-2cos²x)=0
[sinx-1=0⇒sinx=1⇒x=π/2+2πk не удов усл
[2cos²x+cosx-1=0
cosx=t
2t²+t-1=0
D=1+8=9
t1=(-1-3)/4=-1⇒x=π+2πk,k∈z
t2=(-1+3)/4=1/2⇒cosx=1/2⇒x=+-π/3+2πk,k∈z