вправа 22.58 гдз 11 клас алгебра Істер 2019
Вправа 22.58
Умова:
Розв'яжіть рівняння:
\begin{equation} 1)\frac{4}{ctg(2x-\frac{\Pi }{8})}=-4; \end{equation} \begin{equation} 2)\frac{16}{sin(2x+\frac{\Pi }{6})}=-8. \end{equation}
Розв'яжіть рівняння:
\begin{equation} 1)\frac{4}{ctg(2x-\frac{\Pi }{8})}=-4; \end{equation} \begin{equation} 2)\frac{16}{sin(2x+\frac{\Pi }{6})}=-8. \end{equation}
Відповідь ГДЗ:
\begin{equation} 1)\frac{4}{ctg(2x-\frac{\Pi }{8})}=-4 \end{equation} ОДЗ: \begin{equation} {ctg(2x-\frac{\Pi }{8})}\neq 0 \end{equation} \begin{equation} 2x-\frac{\Pi }{8}\neq \frac{\Pi }{2}+k\Pi \end{equation} \begin{equation} 2x\neq \frac{\Pi }{8}+\frac{\Pi }{2}+k\Pi \end{equation} \begin{equation} 2x\neq \frac{5}{8}\Pi+k\Pi \end{equation} \begin{equation} x\neq \frac{5}{16}\Pi+\frac{k\Pi}{2},k\in Z \end{equation} \begin{equation} 4=-4\cdot ctg(2x-\frac{\Pi}{8})\div4 \end{equation} \begin{equation} ctg(2x-\frac{\Pi}{8})=-1 \end{equation} \begin{equation} 2x-\frac{\Pi}{8}=\frac{3\Pi}{4}+k\Pi \end{equation} \begin{equation} 2x=\frac{3\Pi}{4}+\frac{\Pi}{8}+k\Pi \end{equation} \begin{equation} 2x=\frac{7\Pi}{8}+k\Pi \end{equation} \begin{equation} 2x=\frac{7\Pi}{16}+\frac{k\Pi}{2},k\in Z \end{equation} \begin{equation} 2)\frac{16}{sin(2x+\frac{\Pi }{6})}=-8 \end{equation} ОДЗ: \begin{equation} sin(2x+\frac{\Pi}{6})+1\neq 0 \end{equation} \begin{equation} sin(2x+\frac{\Pi}{6})\neq -1 \end{equation} \begin{equation} 2x+\frac{\Pi}{6}\neq -\frac{\Pi}{2}+2k\Pi \end{equation} \begin{equation} 2x\neq -\frac{\Pi}{2}-\frac{\Pi}{6}+2k\Pi \end{equation} \begin{equation} 2x\neq -\frac{4\Pi}{16}+2k\Pi \end{equation} \begin{equation} x\neq \frac{2}{6}\Pi+k\Pi \end{equation} \begin{equation} x\neq \frac{1}{3}\Pi+k\Pi,k\in Z \end{equation} \begin{equation} 8\cdot (sin(2x+\frac{\Pi}{6}+1))=16\div 8 \end{equation} \begin{equation} sin(2x+\frac{\Pi}{6})+1=2 \end{equation} \begin{equation} sin(2x+\frac{\Pi}{6})=1 \end{equation} \begin{equation} 2x+\frac{\Pi}{6}=\frac{\Pi}{2}+2\Pi k \end{equation} \begin{equation} 2x=\frac{\Pi}{2}-\frac{\Pi}{6}+2\Pi k \end{equation} \begin{equation} 2x=\frac{2\Pi}{6}+2\Pi k \end{equation} \begin{equation} 2x=\frac{1}{3}\Pi +2\Pi k \end{equation} \begin{equation} x=\frac{1}{6}\Pi +\Pi k \end{equation}
\begin{equation} 1)\frac{4}{ctg(2x-\frac{\Pi }{8})}=-4 \end{equation} ОДЗ: \begin{equation} {ctg(2x-\frac{\Pi }{8})}\neq 0 \end{equation} \begin{equation} 2x-\frac{\Pi }{8}\neq \frac{\Pi }{2}+k\Pi \end{equation} \begin{equation} 2x\neq \frac{\Pi }{8}+\frac{\Pi }{2}+k\Pi \end{equation} \begin{equation} 2x\neq \frac{5}{8}\Pi+k\Pi \end{equation} \begin{equation} x\neq \frac{5}{16}\Pi+\frac{k\Pi}{2},k\in Z \end{equation} \begin{equation} 4=-4\cdot ctg(2x-\frac{\Pi}{8})\div4 \end{equation} \begin{equation} ctg(2x-\frac{\Pi}{8})=-1 \end{equation} \begin{equation} 2x-\frac{\Pi}{8}=\frac{3\Pi}{4}+k\Pi \end{equation} \begin{equation} 2x=\frac{3\Pi}{4}+\frac{\Pi}{8}+k\Pi \end{equation} \begin{equation} 2x=\frac{7\Pi}{8}+k\Pi \end{equation} \begin{equation} 2x=\frac{7\Pi}{16}+\frac{k\Pi}{2},k\in Z \end{equation} \begin{equation} 2)\frac{16}{sin(2x+\frac{\Pi }{6})}=-8 \end{equation} ОДЗ: \begin{equation} sin(2x+\frac{\Pi}{6})+1\neq 0 \end{equation} \begin{equation} sin(2x+\frac{\Pi}{6})\neq -1 \end{equation} \begin{equation} 2x+\frac{\Pi}{6}\neq -\frac{\Pi}{2}+2k\Pi \end{equation} \begin{equation} 2x\neq -\frac{\Pi}{2}-\frac{\Pi}{6}+2k\Pi \end{equation} \begin{equation} 2x\neq -\frac{4\Pi}{16}+2k\Pi \end{equation} \begin{equation} x\neq \frac{2}{6}\Pi+k\Pi \end{equation} \begin{equation} x\neq \frac{1}{3}\Pi+k\Pi,k\in Z \end{equation} \begin{equation} 8\cdot (sin(2x+\frac{\Pi}{6}+1))=16\div 8 \end{equation} \begin{equation} sin(2x+\frac{\Pi}{6})+1=2 \end{equation} \begin{equation} sin(2x+\frac{\Pi}{6})=1 \end{equation} \begin{equation} 2x+\frac{\Pi}{6}=\frac{\Pi}{2}+2\Pi k \end{equation} \begin{equation} 2x=\frac{\Pi}{2}-\frac{\Pi}{6}+2\Pi k \end{equation} \begin{equation} 2x=\frac{2\Pi}{6}+2\Pi k \end{equation} \begin{equation} 2x=\frac{1}{3}\Pi +2\Pi k \end{equation} \begin{equation} x=\frac{1}{6}\Pi +\Pi k \end{equation}