вправа 23.28 гдз 11 клас алгебра Істер 2019

Відповідь ГДЗ:

1)
\begin{equation} sinx<\frac{\sqrt{3}}{2} \end{equation} \begin{equation} y=sinx,y=\frac{\sqrt{3}}{2} \end{equation} \begin{equation} arcsin\frac{\sqrt{3}}{2}=\frac{\Pi }{3} \end{equation} \begin{equation} [\frac{\Pi }{3};\frac{\Pi }{3}+2\Pi ] \end{equation} \begin{equation} =>[\frac{\Pi }{3};\frac{7\Pi }{3}] \end{equation} \begin{equation} \frac{2\Pi }{3}+2\Pi n<x<\frac{7\Pi }{3}+2\Pi n,n\in Z \end{equation}
2)
\begin{equation} cosx\geq \frac{\sqrt{2}}{2} \end{equation} \begin{equation} arccosx\frac{\sqrt{2}}{2}=\frac{\Pi }{4} \end{equation} \begin{equation} y=cosx,y=\frac{\sqrt{2}}{2} \end{equation} \begin{equation} [-\frac{\Pi }{2}+2\Pi ;\frac{\Pi }{4}+2\Pi ] \end{equation} \begin{equation} -\frac{\Pi }{2}+2\Pi n\leq x\leq \frac{\Pi }{4}+2\Pi n \end{equation}
3)
\begin{equation} tgx<-\frac{1}{\sqrt{3}} \end{equation} \begin{equation} arctgx(-\frac{1}{\sqrt{3}})=\frac{5\Pi }{6} \end{equation} \begin{equation} y=tgx \end{equation} \begin{equation} y=-\frac{1}{\sqrt{3}} \end{equation} \begin{equation} [-\frac{\Pi }{2};\frac{\Pi }{2}] \end{equation} \begin{equation} -\frac{\Pi }{2}+\Pi n;-\frac{\Pi }{6}+\Pi n \end{equation} \begin{equation} -\frac{\Pi }{2}+\Pi n<-\frac{5\Pi }{6}+\Pi n,n\in Z. \end{equation}
4)
\begin{equation} ctgx<\sqrt{3} \end{equation} \begin{equation} arctgx\sqrt{3}=\frac{\Pi }{6},(0;\Pi ) \end{equation} \begin{equation} y=ctgx,y=\sqrt{3} \end{equation} \begin{equation} \frac{5\Pi }{6}+n\Pi <x<\frac{\Pi }{6}+\Pi n,n\in Z. \end{equation}