Тригонометрическое уравнение:(tgx+tg2x)\(1-tgx*tg2x)=-1Если можно,то с решением пожалуйста

[latex]tgx=y[/latex] [latex]tg2x= \frac{2tgx}{1-tg^{2}x}= \frac{2y}{1-y^{2}}[/latex] [latex]tgx+tg2x=y+ \frac{2y}{1-y^{2}}= \frac{y-y^{3}+2y}{1-y^{2}}= \frac{-y^{3}+3y}{1-y^{2}}[/latex] [latex]1-tgx*tg2x=1-y* \frac{2y}{1-y^{2}}= \frac{1-y^{2}-2y^{2}}{1-y^{2}}= \frac{-3y^{2}+1}{1-y^{2}}[/latex] [latex]\frac{-y^{3}+3y}{1-y^{2}}: \frac{-3y^{2}+1}{1-y^{2}}=-1[/latex] [latex] \left \{ {{\frac{-y^{3}+3y}{-3y^{2}+1}=-1} \atop {1-y^{2} \neq 0}} \right. [/latex] [latex] \left \{ {{-y^{3}+3y=3y^{2}-1; 3y^{2}-1 \neq 0} \atop {y^{2} \neq 1}} \right. [/latex] [latex]y^{3}+3y^{2}-3y-1=0[/latex] [latex](y-1)(y^{2}+y+1)+3y(y-1)=0[/latex] [latex](y-1)(y^{2}+y+1+3y)=0[/latex] [latex](y-1)(y^{2}+4y+1)=0[/latex] [latex]y=1[/latex]  или   [latex]y^{2}+4y+1=0[/latex] не удовл.    или   [latex]\frac{D}{4}=2^{2}-1=3[/latex] [latex]y_{1,2}=-2 \pm \sqrt{3}[/latex] [latex]tgx =-2 \pm \sqrt{3}[/latex] [latex]x= arctg(-2 \pm \sqrt{3})+ \pi n,n \in Z[/latex]