Question

$sin3xcosx-sinxcos3x=\frac{\sqrt{2} }{2}$ Нужна помощ

Answer 1

$Sin3xCosx-SinxCos3x=\frac{\sqrt{2} }{2}\\\\Sin(3x-x)=\frac{\sqrt{2} }{2}\\\\Sin2x=\frac{\sqrt{2} }{2}\\\\2x=(-1)^{n}arcSin\frac{\sqrt{2} }{2}+\pi n,n e z\\\\2x=(-1)^{n}\frac{\pi }{4}+\pi n,n e z\\\\x=(-1)^{n}\frac{\pi }{8} +\frac{\pi n }{2},n e z$