Tgx - 1/tg2x = 2 - 1/sin2x

[latex] \frac{sinx}{cosx}- \frac{cos2x}{sin2x}=2-\frac{1}{sin2x}[/latex] [latex] \frac{sinx}{cosx}- \frac{cos2x}{2sinx*cosx}+\frac{1}{2sinx*cosx}-2=0[/latex] [latex] \frac{2sin^{2}x-cos2x+1-4sinx*cosx}{2sinx*cosx}=0[/latex] [latex] \frac{2sin^{2}x-(1-2sin^{2}x)+1-4sinx*cosx}{2sinx*cosx}=0[/latex] [latex] \frac{2sin^{2}x-1+2sin^{2}x+1-4sinx*cosx}{2sinx*cosx}=0[/latex] [latex] \frac{2sin^{2}x-2sinx*cosx}{sinx*cosx}=0[/latex] [latex] \frac{2sinx*(sinx-cosx)}{sinx*cosx}=0[/latex] [latex] \frac{2*(sinx-cosx)}{cosx}=0[/latex] [latex] \left \{ {{sinx-cosx=0} \atop {sinx \neq 0,cosx \neq 0}} \right. [/latex] [latex]sinx=cosx[/latex] [latex]tgx=1[/latex] [latex]x= \frac{ \pi }{4}+ \pi k[/latex], k∈Z [latex]sinx \neq 0[/latex] [latex]x \neq \pi k[/latex], k∈Z [latex]cosx \neq 0[/latex] [latex]x \neq\frac{ \pi }{2}+ \pi k[/latex], k∈Z Ответ: pi/4 + pi*k