ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 727

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\[1)\sin{2x} = - \frac{1}{2}\]

\[2x = ( - 1)^{n + 1} \bullet \arcsin\frac{1}{2} + \pi n\]

\[2x = ( - 1)^{n + 1} \bullet \frac{\pi}{6} + \pi n\]

\[x = \frac{1}{2} \bullet \left( ( - 1)^{n + 1} \bullet \frac{\pi}{6} + \pi n \right)\]

\[x = ( - 1)^{n + 1} \bullet \frac{\pi}{12} + \frac{\text{πn}}{2};\]

\[\left\lbrack - \frac{3\pi}{2};\ \pi \right\rbrack:\]

\[x_{1} = \frac{\pi}{12} - \frac{3\pi}{2} = - \frac{17\pi}{12};\]

\[x_{2} = - \frac{\pi}{12} - \pi = - \frac{13\pi}{12};\]

\[x_{3} = \frac{\pi}{12} - \frac{\pi}{2} = - \frac{5\pi}{12};\]

\[x_{4} = - \frac{\pi}{12};\]

\[x_{5} = \frac{\pi}{12} + \frac{\pi}{2} = \frac{7\pi}{12};\]

\[x_{6} = - \frac{\pi}{12} + \pi = \frac{11\pi}{12}.\]

\[2)\sin{3x} = \frac{\sqrt{3}}{2}\]

\[3x = ( - 1)^{n} \bullet \arcsin\frac{\sqrt{3}}{2} + \pi n\]

\[3x = ( - 1)^{n} \bullet \frac{\pi}{3} + \pi n\]

\[x = \frac{1}{3} \bullet \left( ( - 1)^{n} \bullet \frac{\pi}{3} + \pi n \right)\]

\[x = ( - 1)^{n} \bullet \frac{\pi}{9} + \frac{\text{πn}}{3};\]

\[\left\lbrack - \frac{3\pi}{2};\ \pi \right\rbrack:\]

\[x_{1} = \frac{\pi}{9} - \frac{4\pi}{3} = - \frac{11\pi}{9};\]

\[x_{2} = - \frac{\pi}{9} - \pi = - \frac{10\pi}{9};\]

\[x_{3} = \frac{\pi}{9} - \frac{2\pi}{3} = - \frac{5\pi}{9};\]

\[x_{4} = - \frac{\pi}{9} - \frac{\pi}{3} = - \frac{4\pi}{9};\]

\[x_{5} = \frac{\pi}{9};\]

\[x_{6} = - \frac{\pi}{9} + \frac{\pi}{3} = \frac{2\pi}{9};\]

\[x_{7} = \frac{\pi}{9} + \frac{2\pi}{3} = \frac{7\pi}{9};\]

\[x_{8} = - \frac{\pi}{9} + \pi = \frac{8\pi}{9}.\]