Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 620

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\[1)\sin^{2}x = \frac{1}{4}\]

\[\frac{1 - \cos{2x}}{2} = \frac{1}{4}\]

\[4\left( 1 - \cos{2x} \right) = 2\]

\[4 - 4\cos{2x} = 2\]

\[4\cos{2x} = 4 - 2\]

\[\cos{2x} = \frac{1}{2}\]

\[2x = \pm \arccos\frac{1}{2} + 2\pi n\]

\[2x = \pm \frac{\pi}{3} + 2\pi n\]

\[x = \frac{1}{2} \bullet \left( \pm \frac{\pi}{3} + 2\pi n \right)\]

\[x = \pm \frac{\pi}{6} + \pi n.\]

\[Ответ:\ \pm \frac{\pi}{6} + \pi n.\]

\[2)\cos^{2}x = \frac{1}{2}\]

\[\frac{1 + \cos{2x}}{2} = \frac{1}{2}\]

\[1 + \cos{2x} = 1\]

\[\cos{2x} = 0\]

\[2x = \arccos 0 + \pi n\]

\[2x = \frac{\pi}{2} + \pi n\]

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

\[x = \frac{\pi}{4} + \frac{\text{πn}}{2}.\]

\[Ответ:\ \ \frac{\pi}{4} + \frac{\text{πn}}{2}.\]

\[3)\ 2\sin^{2}x + \sin x - 1 = 0\]

\[y = \sin x:\]

\[2y^{2} + y - 1 = 0\]

\[D = 1 + 8 = 9\]

\[y_{1} = \frac{- 1 - 3}{2 \bullet 2} = - 1;\]

\[y_{2} = \frac{- 1 + 3}{2 \bullet 2} = \frac{1}{2}.\]

\[\sin x = - 1\]

\[x = - \arcsin 1 + 2\pi n\]

\[x = - \frac{\pi}{2} + 2\pi n.\]

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

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

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

\[Ответ:\ \]

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

\[4)\ 2\cos^{2}x + \cos x - 6 = 0\]

\[y = \cos x:\]

\[2y^{2} + y - 6 = 0\]

\[D = 1 + 48 = 49\]

\[y_{1} = \frac{- 1 - 7}{2 \bullet 2} = - 2;\text{\ \ }\]

\[y_{2} = \frac{- 1 + 7}{2 \bullet 2} = \frac{6}{4};\]

\[|y| > 1 - корней\ нет.\]

\[Ответ:\ \ корней\ нет.\]