Sin5xcosx+cos5xsinx=-корень из3/2

[latex]sin5xcosx+sinxcos5x= \frac{\sqrt{3} }{2} \\ \frac{sin(5x-x)+sin(5x+x)+sin(5x+x)+sin(x-5x)}{2} =- \frac{ \sqrt{3} }{2} \\ \frac{sin6x+sin4x-sin4x+sin6x}{2} = - \frac{ \sqrt{3} }{2} \\ sin6x=- \frac{ \sqrt{3} }{2} \\ \left \{ {{6x=- \frac{ \pi }{3} +2 \pi n} \atop {6x=- \frac{2 \pi }{3} +2 \pi n}} \right. \\ \left \{ {{x=- \frac{ \pi }{18} }+ \frac{ \pi n}{3} \atop {x=- \frac{ \pi }{9}+ \frac{ \pi n}{3} }} \right. \\ OTBET: - \frac{ \pi }{9}+ \frac{ \pi n}{3} ;- \frac{ \pi }{18} }+\frac{ \pi n}{3}[/latex]