Помогите решить tg x - tg 2x = sin x

[latex] \frac{sinx}{cosx}- \frac{2tgx}{1-tg^{2}x}=sinx[/latex] [latex] \frac{sinx}{cosx}- \frac{ \frac{2sinx}{cosx}}{1- \frac{sin^{2}x}{cos^{2}x}}=sinx[/latex] [latex] \frac{sinx}{cosx}- \frac{ \frac{2sinx*cos^{2}x}{cosx}}{cos^{2}x-sin^{2}x}=sinx[/latex] [latex] \frac{sinx}{cosx}- \frac{2sinx*cosx}{cos^{2}x-sin^{2}x}=sinx[/latex] [latex] \frac{sinx*(cos^{2}x-sin^{2}x)-2sinx*cos^{2}x}{cosx*(cos^{2}x-sin^{2}x)}=sinx[/latex] [latex]sinx*(cos^{2}x-sin^{2}x)-2sinx*cos^{2}x=sinx*cosx*(cos^{2}x-sin^{2}x)[/latex] [latex]sinx*cos^{2}x-sin^{3}x-2sinx*cos^{2}x=sinx*cos^{3}x-sin^{3}x*cosx[/latex] [latex]sinx*cos^{2}x-sin^{3}x-2sinx*cos^{2}x-sinx*cos^{3}x+sin^{3}x*cosx=0[/latex] [latex](-sinx*cos^{2}x-sinx*cos^{3}x)+(sin^{3}x*cosx-sin^{3}x)=0[/latex] [latex]-sinx*cos^{2}x(1+cosx)+sin^{3}x*(cosx-1)=0[/latex] [latex]sinx*(-cos^{2}x*(1+cosx)+sin^{2}x*(cosx-1))=0[/latex] [latex]sinx*(-cos^{2}x-cos^{3}x+sin^{2}x*cosx-sin^{2}x)=0[/latex] [latex]sinx*(-cos^{2}x-cos^{3}x+(1-cos^{2}x)*cosx-1+cos^{2}x)=0[/latex] [latex]sinx*(-cos^{3}x+cosx-cos^{3}x-1)=0[/latex] [latex]sinx*(-2cos^{3}x+cosx-1)=0[/latex] 1) [latex]sinx=0[/latex] [latex]x= \pi k[/latex], k∈Z 2) [latex]-2cos^{3}x+cosx-1=0[/latex] [latex]2cos^{3}x-cosx+1=0[/latex] 2.1) [latex]cosx=-1[/latex] [latex]x= \frac{ \pi }{2}+ \pi k[/latex], k∈Z 2.2) [latex]2cos^{3}x-cosx+1=(cosx+1)(2cos^{2}x-2cosx+1)=0[/latex] [latex]2cos^{2}x-2cosx+1=0[/latex] Замена: [latex]cosx=t[/latex],  t∈[-1;1] [latex]2t^{2}-2t+1=0, D=4-4*1*4=-12\ \textless \ 0[/latex] - нет решений 3) Объединим получившиеся решения в одно: [latex]x= \frac{ \pi }{2}+ \frac{ \pi k }{2}[/latex], k∈Z