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а) Решим уравнение

$\left\{\begin{matrix}{2\sin^{2}x-3\sin x+1=0}\\{\mathrm{tg}x\neq0}\\\end{matrix}\right.\Leftrightarrow\begin{cases}\left[\begin{matrix}{\sin x=\frac{1}{2}}\\{\sin x=1}\end{matrix}\right.\\{\sin x\neq0}\\{\cos x\neq0}\\\end{cases}\Leftrightarrow\sin x=\frac{1}{2}\Leftrightarrow\left[\begin{matrix}{x=\frac{\pi}{6}+2\pi k}\\{x=\frac{5\pi}{6}+2\pi k}\\\end{matrix}\right.,k\in\mathbb{Z}$

б) Отберем корни двойным неравенством

$\begin{gathered} -\pi\leq\frac{\pi}{6}+2\pi k\leq\frac{\pi}{2}\mid:\pi \\ -1\leq\frac{1}{6}+2k\leq\frac{1}{2}\mid-\frac{1}{6} \\ -\frac{7}{6}\leq2k\leq\frac{2}{6}\mid:2 \\ -\frac{7}{12}\leq k\leq\frac{2}{12} \\ k\in Z\Rightarrow k=0\Rightarrow \\ \Rightarrow x_{1}=\frac{\pi}{6} \end{gathered}$

$\begin{gathered} -\pi\leq\frac{5\pi}{6}+2\pi k\leq\frac{\pi}{2}\mid:\pi \\ -1\leq\frac{5}{6}+2k\leq\frac{1}{2}\mid-\frac{5}{6} \\ -\frac{11}{6}\leq2k\leq-\frac{2}{6}\mid:2 \\ -\frac{11}{12}\leq k\leq-\frac{2}{12} \\ k\in\mathbb{Z}\Rightarrow k\in\emptyset \end{gathered}$

48:{"origi

Задание №3775 ЕГЭ по Математике