7f31d90479.js"],""] 32:T790,

а) Решим уравнение

$\\\left[\begin{matrix}\sin x=0\\\sqrt{2}\sin^2x+\cos x-\sqrt{2}=0\\\sin x\leq0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}\sin x=0\\-\sqrt{2}\cos^2x+\cos x=0\\\sin x\leq0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}\sin x=0\\\begin{cases}\cos x\left(1-\sqrt{2}\cos x\right)=0\\\sin x\leq0\end{cases}\end{matrix}\right.\Leftrightarrow\Leftrightarrow\begin{cases}\left[\begin{matrix}\sin x=0\\\cos x=0\\\cos x=\frac{\sqrt{2}}{2}\end{matrix}\right.\\\sin x\leq0\end{cases}\Leftrightarrow\left[\begin{matrix}x=\pi k\\x=-\frac{\pi}{2}+2\pi k\\x=-\frac{\pi}{4}+2\pi k\end{matrix}\right.,k\in\mathbb{Z}\\$

б) Отберем корни двойным неравенством:

$\begin{gathered} 2\pi\leq\pi k\leq{\frac{7\pi}{2}}\mid:\pi \\ 2\leq k\leq\frac{7}{2} \\ k\in\mathbb{Z}\Rightarrow k=2;3\Rightarrow \\ \Rightarrow x_{1}=2\pi;x_{2}=3\pi \end{gathered}$

$\begin{gathered} 2\pi \leq-\frac{\pi}{2}+2\pi k\leq\frac{7\pi}{2}\mid:\pi \\ \text{-} 2\leq-\frac{1}{2}+2k\leq\frac{7}{2}\mid+\frac{1}{2} \\ \frac{5}{2}\leq2k\leq\frac{8}{2}\mid:2 \\ \frac54\leq k\leq\frac84 \\ k\in\mathbb{Z}\Rightarrow k=2\Rightarrow \\ \Rightarrow x_{3}=-\frac{\pi}{2}+4\pi=\frac{7\pi}{2} \end{gathered}$

$\begin{gathered} \text{2} \pi\leq-\frac{\pi}{4}+2\pi k\leq\frac{7\pi}{2}\mid:\pi \\ 2\leq-\frac{1}{4}+2k\leq\frac{7}{2}\mid+\frac{1}{4} \\ \frac{9}{4}\leq2k\leq\frac{15}{4}\mid:2 \\ \frac{9}{8}\leq k\leq\frac{15}{8} \\ k\in\mathbb{Z}\Rightarrow k\in\emptyset \end{gathered}$

Задание №3649 ЕГЭ по Математике