Question

Найдите корни уравнения
sinx=-$frac{ sqrt{2} }{2}$ x∈[-pi;2pi]

Answer 1

$sin(x)=-frac{sqrt{2}}{2}\x=(-1)^n*arcsin(-frac{sqrt{2}}{2})+pi*n, nin Z\x=(-1)^{n+1}*frac{pi}{4}+pi*n, nin Z\\\n=-1;x=(-1)^0*frac{pi}{4}-pi=frac{pi}{4}-frac{4pi}{4}=boxed{-frac{3pi}{4}}\n=0;x=(-1)^1*frac{pi}{4}+pi*0=boxed{-frac{pi}{4}}\n=1;x=(-1)^2*frac{pi}{4}+pi=boxed{frac{3pi}{4}}\n=2;x=(-1)^3*frac{pi}{4}+2pi=-frac{pi}{4}+frac{8pi}{4}=boxed{frac{7pi}{4}}$