Question

НАЙДИТЕ КОРНИ УРАВНЕНИЯ: sin3x=√2/2,[0;2п]

Answer 1

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