Question

Найдите корни уравнения: sin3x=√2/2,[0; 2п]

Answer 1

$sin3x=\frac{\sqrt{2}}{2}\\3x=(-1)^n*arcsin(\frac{\sqrt{2}}{2})+\pi*n,\ n\in Z\\3x=(-1)^n*\frac{\pi}{4}+\pi*n\ n\in Z\\x=(-1)^n*\frac{\pi}{12}+\frac{\pi*n}{3},\ n\in Z\\\\n=0,x=(-1)^0*\frac{\pi}{12}+\frac{\pi*0}{3}=\boxed{\frac{\pi}{12}}\\n=1,x=(-1)^1*\frac{\pi}{12}+\frac{\pi*1}{3}=-\frac{\pi}{12}+\frac{4\pi}{12}=\boxed{\frac{\pi}{4}}\\n=2,x=(-1)^2*\frac{\pi}{12}+\frac{\pi*2}{3}=\frac{\pi}{12}+\frac{8\pi}{12}=\boxed{\frac{3\pi}{4}}$
$n=3,x=(-1)^3*\frac{\pi}{12}+\frac{\pi*3}{3}=-\frac{\pi}{12}+\frac{12\pi}{12}=\boxed{\frac{11\pi}{12}}\\n=4,x=(-1)^4*\frac{\pi}{12}+\frac{\pi*4}{3}=\frac{\pi}{12}+\frac{16\pi}{12}=\boxed{\frac{17\pi}{12}}\\n=5,x=(-1)^5*\frac{\pi}{12}+\frac{\pi*5}{3}=-\frac{\pi}{12}+\frac{20\pi}{12}=\boxed{\frac{19\pi}{12}}$