x+(2+√2)·^2x=1 tenglamaning [0;\ 2π] kesmadagi yechimlari yigʻindisini toping.

$\sin^2x=1-\cos^2x$ almashtirish bilan $(2+\sqrt{2})\cos^2x-\cos x-(1+\sqrt{2})=0$ tenglamaga kelamiz. $\cos x=1$ ildiz ekanini tekshirish oson; Viyet teoremasiga ko‘ra ikkinchi ildiz $\cos x=-\frac{1+\sqrt{2}}{2+\sqrt{2}}=-\frac{\sqrt{2}}{2}$. $[0;\ 2\pi]$ kesmada: $\cos x=1$ dan $x=0$, $x=2\pi$; $\cos x=-\frac{\sqrt{2}}{2}$ dan $x=\frac{3\pi}{4}$, $x=\frac{5\pi}{4}$. Yig‘indi: $0+2\pi+\frac{3\pi}{4}+\frac{5\pi}{4}=4\pi$.

Manba: UZBMB namunaviy test topshiriqlari (2022, mutaxassislik)