Question

Найти кол-во углов принадлежащих 0=

Answer 1

$\sqrt{2}sin4x-(sin7x+sinx)=0\\\sqrt{2}sin4x-(2sin\frac{7x+x}{2}*cos\frac{7x-x}{2})=0\\\sqrt{2}sin4x-2sin4x*cos3x=0\\sin4x(\sqrt{2}-2cos3x)=0\\sin4x=0\\4x=\pi*n,n\in Z\\\boxed{x=\frac{\pi*n}{4},n\in Z}\\\sqrt{2}-2cos3x=0\\cos3x=\frac{\sqrt{2}}{2}\\3x=бarccos\frac{\sqrt{2}}{2}+2\pi*k,k\in Z\\3x=б\frac{\pi}{4}+2\pi*k,k\in Z\\\boxed{x=б\frac{\pi}{12}+\frac{2\pi*k}{3},k\in Z}$

$x=\frac{\pi*n}{4},n\in Z\\n=0,\boxed{x=0}\\n=1,\boxed{x=\frac{\pi}{4}}\\n=2,\boxed{x=\frac{\pi}{2}}$

$x=б\frac{\pi}{12}+\frac{2\pi*k}{3},k\in Z\\x=\frac{\pi}{12}+\frac{2\pi*k}{3},k\in Z\\k=0,\boxed{x=\frac{\pi}{12}}\\x=-\frac{\pi}{12}+\frac{2\pi*k}{3},k\in Z\\k=1,\boxed{x=-\frac{\pi}{12}+\frac{8\pi}{12}=\frac{7\pi}{12}}$

ответ:5