Sin2x = tgx решить уравнение

[tex]sin2x=tgx= \frac{sinx}{cosx} \\\left \{ {{2sinxcos^2x=sinx} \atop {cosx \neq 0}} \right. = \left \{ {{sinx(2cos^2-1)=0} \atop {cosx \neq 0}} \right. = \left \{ {{sinx*cos2x=0} \atop {cosx \neq 0}} \right. \\ sin x =0 \\ x = \pi n \\ cos 2x=0 \\2x = \frac{ \pi }{2} + \pi n \\x = \frac{ \pi }{4} + \frac{ \pi n}{2} [/tex]