Буду благодарен за решение. .

sin12x=1/2*sinx-√3/2*cosx sin12x=sin(x-π/3) {12x=x-π/3+2πn⇒[11x=-π/3+2πn⇒[x=-π/33+2πn/11 {12x=π-x+π/3+2πn⇒[13x=4π/3+2πn⇒[x=4π/39+2πn/13,n∈Z

[latex]sin12x= \frac{1}{2} *sinx- \frac{ \sqrt{3} }{2} *cosx[/latex] [latex]sin12x=sin(x- \frac{ \pi }{3} )[/latex] [latex]12x=x- \frac{ \pi }{3} +2 \pi n =\ \textgreater \ 11x=-\frac{ \pi }{3} +2 \pin =\ \textgreater \ x=- \frac{ \pi }{33} + \frac{2 \pi n}{11} [/latex] [latex]12x= \pi -x+ \frac{ \pi }{3} +2 \pi n=\ \textgreater \ 13x= \frac{4 \pi }{3} +2 \pin=\ \textgreater \ x= \frac{4 \pi }{39} + \frac{2 \pi n}{13}[/latex],n∈Z