вправа 15.5 гдз 10 клас математика Мерзляк Номіровський 2018
Вправа 15.5
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Відповідь ГДЗ:
\begin{equation} 1) \cos \left ( x+\frac{\pi}{6} \right )=\frac{\sqrt{2}}{2}; \end{equation} \begin{equation} x+\frac{\pi}{6}= \pm \arccos \frac{\sqrt{2}}{2}+ \end{equation} \begin{equation} +2 \pi n, n \in Z; \end{equation} \begin{equation} x+\frac{\pi}{6}= \pm \frac{\pi}{4}+ \end{equation} \begin{equation} +2 \pi n, n \in Z; \end{equation} \begin{equation} x= \pm \frac{\pi}{4}-\frac{\pi}{6}+ \end{equation} \begin{equation} +2 \pi n, n \in Z; \end{equation} \begin{equation} 2)\cos \left ( \frac{\pi}{4}-x \right )=-\frac{\sqrt{3}}{2}; \end{equation} \begin{equation} \cos \left ( x-\frac{\pi}{4} \right )= \end{equation} \begin{equation} =\frac{\sqrt{3}}{2}; \end{equation} \begin{equation} x-\frac{\pi}{4}= \end{equation} \begin{equation} =\pm \arccos \frac{\sqrt{3}}{2}+2 \pi n, n \in Z; \end{equation} \begin{equation} x-\frac{\pi}{4}= \end{equation} \begin{equation} =\pm \frac{\pi}{6}+2 \pi n, n \in Z; \end{equation} \begin{equation} x= \pm \frac{\pi}{6}+\frac{\pi}{4}+ \end{equation} \begin{equation} +2 \pi n, n \in Z; \end{equation} \begin{equation} 3)\cos \left ( \frac{x}{6}-2 \right )=-1; \end{equation} \begin{equation} \frac{x}{6}-2= \pi + \pi n, n \in Z; \end{equation} \begin{equation} \frac{x}{6}=\pi +2+2 \pi n, n \in Z; \end{equation} \begin{equation} x=6 \pi +12+12 \pi n, n \in Z; \end{equation} \begin{equation} 4)2 \cos \left ( \frac{\pi}{8}-3x \right )+1=0; \end{equation} \begin{equation} 2 \cos \left (3x-\frac{\pi}{8} \right )=-1; \end{equation} \begin{equation} \cos \left (3x-\frac{\pi}{8} \right )=-\frac{1}{2}; \end{equation} \begin{equation} 3x-\frac{\pi}{8}=\pm \arccos \left ( -\frac{1}{2} \right )+ \end{equation} \begin{equation} +2 \pi n, n \in Z; \end{equation} \begin{equation} 3x-\frac{\pi}{8}=\pm \left (\pi - \arccos \frac{1}{2} \right )+ \end{equation} \begin{equation} +2 \pi n, n \in Z; \end{equation} \begin{equation} 3x-\frac{\pi}{8}= \end{equation} \begin{equation} =\pm \left ( \pi - \frac{\pi}{3} \right )+2 \pi n, n \in Z; \end{equation} \begin{equation} 3x-\frac{\pi}{8}= \end{equation} \begin{equation} = \pm \frac{2 \pi}{3}+2 \pi n, n \in Z; \end{equation} \begin{equation} 3x= \pm \frac{2 \pi}{3}+\frac{\pi}{8}+ \end{equation} \begin{equation} +2 \pi n, n \in Z; \end{equation} \begin{equation} x= \pm \frac{2 \pi}{9}+\frac{\pi}{24}+ \end{equation} \begin{equation} +\frac{2 \pi n}{3}, n \in Z. \end{equation}
\begin{equation} 1) \cos \left ( x+\frac{\pi}{6} \right )=\frac{\sqrt{2}}{2}; \end{equation} \begin{equation} x+\frac{\pi}{6}= \pm \arccos \frac{\sqrt{2}}{2}+ \end{equation} \begin{equation} +2 \pi n, n \in Z; \end{equation} \begin{equation} x+\frac{\pi}{6}= \pm \frac{\pi}{4}+ \end{equation} \begin{equation} +2 \pi n, n \in Z; \end{equation} \begin{equation} x= \pm \frac{\pi}{4}-\frac{\pi}{6}+ \end{equation} \begin{equation} +2 \pi n, n \in Z; \end{equation} \begin{equation} 2)\cos \left ( \frac{\pi}{4}-x \right )=-\frac{\sqrt{3}}{2}; \end{equation} \begin{equation} \cos \left ( x-\frac{\pi}{4} \right )= \end{equation} \begin{equation} =\frac{\sqrt{3}}{2}; \end{equation} \begin{equation} x-\frac{\pi}{4}= \end{equation} \begin{equation} =\pm \arccos \frac{\sqrt{3}}{2}+2 \pi n, n \in Z; \end{equation} \begin{equation} x-\frac{\pi}{4}= \end{equation} \begin{equation} =\pm \frac{\pi}{6}+2 \pi n, n \in Z; \end{equation} \begin{equation} x= \pm \frac{\pi}{6}+\frac{\pi}{4}+ \end{equation} \begin{equation} +2 \pi n, n \in Z; \end{equation} \begin{equation} 3)\cos \left ( \frac{x}{6}-2 \right )=-1; \end{equation} \begin{equation} \frac{x}{6}-2= \pi + \pi n, n \in Z; \end{equation} \begin{equation} \frac{x}{6}=\pi +2+2 \pi n, n \in Z; \end{equation} \begin{equation} x=6 \pi +12+12 \pi n, n \in Z; \end{equation} \begin{equation} 4)2 \cos \left ( \frac{\pi}{8}-3x \right )+1=0; \end{equation} \begin{equation} 2 \cos \left (3x-\frac{\pi}{8} \right )=-1; \end{equation} \begin{equation} \cos \left (3x-\frac{\pi}{8} \right )=-\frac{1}{2}; \end{equation} \begin{equation} 3x-\frac{\pi}{8}=\pm \arccos \left ( -\frac{1}{2} \right )+ \end{equation} \begin{equation} +2 \pi n, n \in Z; \end{equation} \begin{equation} 3x-\frac{\pi}{8}=\pm \left (\pi - \arccos \frac{1}{2} \right )+ \end{equation} \begin{equation} +2 \pi n, n \in Z; \end{equation} \begin{equation} 3x-\frac{\pi}{8}= \end{equation} \begin{equation} =\pm \left ( \pi - \frac{\pi}{3} \right )+2 \pi n, n \in Z; \end{equation} \begin{equation} 3x-\frac{\pi}{8}= \end{equation} \begin{equation} = \pm \frac{2 \pi}{3}+2 \pi n, n \in Z; \end{equation} \begin{equation} 3x= \pm \frac{2 \pi}{3}+\frac{\pi}{8}+ \end{equation} \begin{equation} +2 \pi n, n \in Z; \end{equation} \begin{equation} x= \pm \frac{2 \pi}{9}+\frac{\pi}{24}+ \end{equation} \begin{equation} +\frac{2 \pi n}{3}, n \in Z. \end{equation}