Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 1258

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\[1)\ \frac{2\cos x + \sin^{2}x}{ctg\ x - \sin{2x}} = tg\ 2x\]

\[- \frac{\sin x\left( 3\cos^{2}x - \cos x - 1 \right)}{2\cos^{2}x - \cos x} = 0\]

\[3\cos^{2}x - 2\cos x - 1 = 0\]

\[Пусть\cos x = y:\]

\[3y^{2} - 2y - 1 = 0\]

\[D_{1} = 1 + 3 = 4\]

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

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

\[\cos x = - \frac{1}{3}\]

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

\[Ответ:\ \pm \frac{1}{3}\arccos\left( - \frac{1}{3} \right) + 2\pi n.\]

\[2)\ \frac{ctg\ x - tg\ x}{\cos x + 3\cos{2x}} = ctg\ 2x\]

\[6\cos x - 5 = 0\]

\[\cos x = \frac{5}{6}\]

\[x = \pm \arccos\frac{5}{6} + 2\pi n.\]

\[\cos x = 1\]

\[не\ подходит.\]

\[2\cos^{2}x - 1 = 0\]

\[2 \cdot \frac{1 - \cos{2x}}{2} - 1 = 0\]

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

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

\[x = \frac{\pi}{4} + \pi n.\]

\[Ответ:\ \pm \arccos\frac{5}{6} + 2\pi n;\]

\[\text{\ \ \ }\frac{\pi}{4} + \pi n.\]

\[3)\ \frac{\cos{3x} - \sin x}{\cos{5x} - \sin{3x}} = 1\]

\[\frac{\cos{3x} - \cos\left( \frac{\pi}{2} - x \right)}{\cos{5x} - \cos\left( \frac{\pi}{2} - 3x \right)} = 1\]

\[\frac{- 2\sin\left( 2x - \frac{\pi}{4} \right)\sin\left( x + \frac{\pi}{4} \right)}{- 2\sin\left( x + \frac{\pi}{4} \right)\sin\left( 4x - \frac{\pi}{4} \right)} = 1\]

\[\sin\left( 2x - \frac{\pi}{4} \right) - \sin\left( 4x - \frac{\pi}{4} \right) = 0\]

\[\sin x\cos\left( 3x - \frac{\pi}{4} \right) = 0\]

\[\sin x = 0\]

\[x = \pi n.\]

\[\cos\left( 3x - \frac{\pi}{4} \right) = 0\]

\[3x - \frac{\pi}{4} = \frac{\pi}{2} + \pi n\]

\[3x = \frac{3\pi}{4} + \pi n\]

\[x = \frac{\pi}{4} + \frac{\text{πn}}{3}.\]

\[Ответ:\ \ \pi n;\ \ \frac{\pi}{4} + \frac{\text{πn}}{3}.\]

\[4)\ \frac{\cos{3x} + \sin{5x}}{\cos x + \sin{3x}} = - 1\]

\[\frac{\cos{3x} + \cos\left( \frac{\pi}{2} - 5x \right)}{\cos x + \cos\left( \frac{\pi}{2} - 3x \right)} = - 1\]

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

\[\cos\left( 4x - \frac{\pi}{4} \right) + \cos\left( 2x - \frac{\pi}{4} \right) = 0\]

\[\cos\left( 3x - \frac{\pi}{4} \right)\cos x = 0\]

\[\cos x = 0\]

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

\[\cos\left( 3x - \frac{\pi}{4} \right) = 0\]

\[3x - \frac{\pi}{4} = \frac{\pi}{2} + \pi n\]

\[3x = \frac{3\pi}{4} + \pi n\]

\[x = \frac{\pi}{4} + \frac{\text{πn}}{3}.\]

\[Ответ:x = \frac{\pi}{2} + \pi n;\ \ x = \frac{\pi}{4} + \frac{\text{πn}}{3}.\]