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а) Решим уравнение:

$\cos^{2}x(2\cos x-1)+(2\cos x-1)=0$ $(2\cos x-1)(\cos^{2}x+1)=0$ $\left[\begin{matrix}2\cos x-1=0\\\cos^{2}x+1=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}\cos x=\frac{1}{2}\\x\in\emptyset\end{matrix}\right.\Leftrightarrow x=\pm\frac{\pi}{3}+2\pi k,k\in Z$

б) Отберем корни с помощью двойного неравенства:

$\begin{aligned} 2\pi\le\frac\pi3+2\pi k\le\frac{7\pi}2\mid:\pi\\ 2\le\frac13+2k\le\frac72\mid-\frac13\\ \frac53\le2k\le\frac{19}6\mid:2\\ \frac56\le k\le\frac{19}{12}\\ \frac{10}{12}\le k\le\frac{19}{12}\\ k=1\\ x_1=\frac\pi3+2\pi=\frac{7\pi}3 \end{aligned}$

$\begin{aligned} 2\pi\le-\frac\pi3+2\pi k\le\frac{7\pi}2\mid:\pi\\ 2\le-\frac13+2k\le\frac72\mid+\frac13\\ \frac73\le2k\le\frac{23}6\mid:2\\ \frac76\le k\le\frac{23}{12}\\ \frac{14}{12}\le k\le\frac{23}{12}\\ k\in\emptyset \end{aligned}$

48:{"original_url":"https://storage.yandexcloud.net/egehub/public/776805/xROJAaiT4TVwNm1CfpuKJt

Задание №3708 ЕГЭ по Математике