вправа 16.9 гдз 10 клас математика Мерзляк Номіровський 2018
Вправа 16.9
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Відповідь ГДЗ:
\begin{equation} 1)\sin \left ( x-\frac{\pi}{6} \right )=\frac{\sqrt{2}}{2}; \end{equation} \begin{equation} x-\frac{\pi}{6}=(-1)^{n} \arcsin \frac{\sqrt{2}}{2}+ \end{equation} \begin{equation} + \pi n, n \in Z; \end{equation} \begin{equation} x=(-1)^{n}\frac{\pi}{4}+\frac{\pi}{6}+ \end{equation} \begin{equation} +\pi n, n \in Z; \end{equation} \begin{equation} 2)\sin \left ( \frac{\pi}{8}-x \right )=\frac{\sqrt{3}}{2}; \end{equation} \begin{equation} \sin \left ( x-\frac{\pi}{8} \right )=-\frac{\sqrt{3}}{2}; \end{equation} \begin{equation} x-\frac{\pi}{8}= \end{equation} \begin{equation} =(-1)^{n} \arcsin \left ( -\frac{\sqrt{3}}{2} \right )+ \end{equation} \begin{equation} +\pi n, n \in Z; \end{equation} \begin{equation} x=(-1)^{n+1}\frac{\pi}{3}+\frac{\pi}{8}+ \end{equation} \begin{equation} +\pi n, n \in Z; \end{equation} \begin{equation} 3)\sin \left ( \frac{x}{3}+1 \right )=-1; \end{equation} \begin{equation} \frac{x}{3}+1=-\frac{\pi}{2}+2\pi n, n \in Z; \end{equation} \begin{equation} \frac{x}{3}=-\frac{\pi}{2}-1+2\pi n, n \in Z; \end{equation} \begin{equation} x=-\frac{3 \pi}{2}-3+6 \pi n, n \in Z; \end{equation} \begin{equation} 4) \sin \left ( \frac{\pi}{12}-3x \right )=\frac{1}{\sqrt{2}}; \end{equation} \begin{equation} \sin \left ( 3x-\frac{\pi}{12} \right )= \end{equation} \begin{equation} =-\frac{\sqrt{2}}{2}; \end{equation} \begin{equation} 3x-\frac{\pi}{12}= \end{equation} \begin{equation} =(-1)^{n} \arcsin \left ( -\frac{\sqrt{2}}{2} \right )+ \end{equation} \begin{equation} +\pi n, n \in Z; \end{equation} \begin{equation} 3x=(-1)^{n+1}\frac{\pi}{4}+\frac{\pi}{12}+ \end{equation} \begin{equation} +\pi n, n \in Z; \end{equation} \begin{equation} x=(-1)^{n+1}\frac{\pi}{12}+\frac{\pi}{36}+ \end{equation} \begin{equation} +\frac{\pi n}{3}, n \in Z. \end{equation}
\begin{equation} 1)\sin \left ( x-\frac{\pi}{6} \right )=\frac{\sqrt{2}}{2}; \end{equation} \begin{equation} x-\frac{\pi}{6}=(-1)^{n} \arcsin \frac{\sqrt{2}}{2}+ \end{equation} \begin{equation} + \pi n, n \in Z; \end{equation} \begin{equation} x=(-1)^{n}\frac{\pi}{4}+\frac{\pi}{6}+ \end{equation} \begin{equation} +\pi n, n \in Z; \end{equation} \begin{equation} 2)\sin \left ( \frac{\pi}{8}-x \right )=\frac{\sqrt{3}}{2}; \end{equation} \begin{equation} \sin \left ( x-\frac{\pi}{8} \right )=-\frac{\sqrt{3}}{2}; \end{equation} \begin{equation} x-\frac{\pi}{8}= \end{equation} \begin{equation} =(-1)^{n} \arcsin \left ( -\frac{\sqrt{3}}{2} \right )+ \end{equation} \begin{equation} +\pi n, n \in Z; \end{equation} \begin{equation} x=(-1)^{n+1}\frac{\pi}{3}+\frac{\pi}{8}+ \end{equation} \begin{equation} +\pi n, n \in Z; \end{equation} \begin{equation} 3)\sin \left ( \frac{x}{3}+1 \right )=-1; \end{equation} \begin{equation} \frac{x}{3}+1=-\frac{\pi}{2}+2\pi n, n \in Z; \end{equation} \begin{equation} \frac{x}{3}=-\frac{\pi}{2}-1+2\pi n, n \in Z; \end{equation} \begin{equation} x=-\frac{3 \pi}{2}-3+6 \pi n, n \in Z; \end{equation} \begin{equation} 4) \sin \left ( \frac{\pi}{12}-3x \right )=\frac{1}{\sqrt{2}}; \end{equation} \begin{equation} \sin \left ( 3x-\frac{\pi}{12} \right )= \end{equation} \begin{equation} =-\frac{\sqrt{2}}{2}; \end{equation} \begin{equation} 3x-\frac{\pi}{12}= \end{equation} \begin{equation} =(-1)^{n} \arcsin \left ( -\frac{\sqrt{2}}{2} \right )+ \end{equation} \begin{equation} +\pi n, n \in Z; \end{equation} \begin{equation} 3x=(-1)^{n+1}\frac{\pi}{4}+\frac{\pi}{12}+ \end{equation} \begin{equation} +\pi n, n \in Z; \end{equation} \begin{equation} x=(-1)^{n+1}\frac{\pi}{12}+\frac{\pi}{36}+ \end{equation} \begin{equation} +\frac{\pi n}{3}, n \in Z. \end{equation}