July 2022 1 3 Report
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Участник ЗнанийУчастник Знаний
sin12x= \frac{1}{2} *sinx- \frac{ \sqrt{3} }{2} *cosx
sin12x=sin(x- \frac{ \pi }{3} )
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}
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},n∈Z
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