Question

Найдите корни уравнения sinx=- x∈[-pi;2pi]

Answer 1

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