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МатематикаГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1257
Задание 1257
\(\boxed{\mathbf{1257}\mathbf{.}}\)
\[1)\sin{3x} + \left| \sin x \right| = \sin{2x}\]
\[\left| \sin x \right| = - \sin x \bullet\]
\[\bullet \left( 4\cos^{2}x - 2\cos x - 1 \right)\]
\[1)\sin x > 0:\]
\[4\cos^{2}x - 2\cos x - 1 = - 1\]
\[4\cos^{2}x - 2\cos x = 0\]
\[2\cos x\left( 2\cos x - 1 \right) = 0\]
\[\cos x = 0\]
\[x = \frac{\pi}{2} + \pi n.\]
\[2\cos x - 1 = 0\]
\[\cos x = \frac{1}{2}\]
\[x = \pm \frac{\pi}{3} + 2\pi n.\]
\[Нам\ подходят\ корни:\]
\[x = \frac{\pi}{2} + 2\pi n;\ \ x = \frac{\pi}{3} + 2\pi n.\]
\[2)\sin x < 0:\]
\[4\cos^{2}x - 2\cos x - 1 = 1\]
\[4\cos^{2}x - 2\cos x - 2 = 0\]
\[2\cos^{2}x - \cos x - 1 = 0\]
\[Пусть\cos x = y:\]
\[2y^{2} - y - 1 = 0\]
\[D = 1 + 8 = 9\]
\[y_{1} = \frac{1 + 3}{4} = 1;\ \ \ \]
\[y_{2} = \frac{1 - 3}{4} = - \frac{1}{2}.\]
\[\cos x = 1\]
\[x = \pi + \pi n.\]
\[\cos x = - \frac{1}{2}\]
\[x = \pm \frac{2\pi}{3} + 2\pi n.\]
\[Нам\ подходит:\]
\[x = - \frac{2\pi}{3} + 2\pi n.\]
\[\sin x = 0\]
\[x = \pi n.\]
\[Ответ:\ \pi n;\ \ \frac{\pi}{3} + \pi n;\ \ \ \frac{\pi}{2} + 2\pi n.\]
\[2)\cos{3x} + \left| \cos x \right| = \sin{2x}\]
\[\left| \cos x \right| = \sin{2x} - \cos{3x}\]
\[\left| \cos x \right| = \cos x \bullet\]
\[\bullet \left( - 4\cos^{2}x + 2\sin x + 3 \right)\]
\[\left| \cos x \right| = \cos x \bullet\]
\[\bullet (4\sin^{2}x + 2\sin x - 1)\]
\[1)\cos x = 0:\]
\[x = \frac{\pi}{2} + \pi n.\]
\[2)\cos x > 0:\]
\[4\sin^{2}x + 2\sin x - 2 = 0\]
\[Пусть\sin x = y:\]
\[2y^{2} + y - 1 = 0\]
\[D = 1 + 8 = 9\]
\[y_{1} = \frac{- 1 + 3}{4} = \frac{1}{2};\ \ \ \]
\[\ y_{2} = \frac{- 1 - 3}{4} = - 1.\]
\[\sin x = - 1\]
\[x = - \frac{\pi}{2} + \pi n.\]
\[\sin x = - \frac{1}{2}\]
\[x = ( - 1)^{n} \cdot \frac{\pi}{6} + \pi n.\]
\[Нам\ подходит\ корень:\]
\[x = \frac{\pi}{6} + 2\pi n.\]
\[3)\cos x < 0:\]
\[4\sin^{2}x + 2\sin x = 0\]
\[2\sin x\left( 2\sin x + 1 \right) = 0\]
\[\sin x = 0\]
\[x = \pi n.\]
\[\sin x = - \frac{1}{2}\]
\[x = ( - 1)^{n} \cdot \frac{\pi}{6} + \pi n.\]
\[Нам\ подходит\ корень:\]
\[x = \pi + 2\pi n.\]
\[Ответ:\ \ x = \pi + 2\pi n;\ \ \]
\[x = \frac{\pi}{2} + \pi n;\ \ x = \frac{\pi}{6} + 2\pi n.\]
