Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1564

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\[1)\ \frac{\sin{3x}}{\cos{2x}} + \frac{\cos{3x}}{\sin{2x}} = \frac{2}{\sin{3x}}\]

\[\frac{\sin{3x} \bullet \sin{2x} + \cos{3x} \bullet \cos{2x}}{\sin{2x} \bullet \cos{2x}} = \frac{2}{\sin{3x}}\]

\[\frac{\cos(3x - 2x)}{\sin{2x} \bullet \cos{2x}} - \frac{2}{\sin{3x}} = 0\]

\[\frac{\cos x \bullet \sin{3x} - 2\sin{2x} \bullet \cos{2x}}{\sin{2x} \bullet \cos{2x} \bullet \sin{3x}} = 0\]

\[\frac{\cos x \bullet \sin{3x} - \sin{4x}}{\frac{1}{2}\sin{4x} \bullet \sin{3x}} = 0\]

\[\frac{\frac{1}{2}\left( \sin{4x} + \sin{2x} \right) - \sin{4x}}{\frac{1}{2}\sin{4x} \bullet \sin{3x}} = 0\]

\[\frac{\sin{4x} + \sin{2x} - 2\sin{4x}}{\sin{4x} \bullet \sin{3x}} = 0\]

\[\frac{\sin{2x} - \sin{4x}}{\sin{4x} \bullet \sin{3x}} = 0\]

\[\frac{2 \bullet \sin\frac{2x - 4x}{2} \bullet \cos\frac{2x + 4x}{2}}{\sin{4x} \bullet \sin{3x}} = 0\]

\[\frac{- 2 \bullet \sin x \bullet \cos{3x}}{\sin{4x} \bullet \sin{3x}} = 0\]

\[1)\ \sin x = 0\]

\[x = \arcsin 0 + \pi n = \pi n.\]

\[2)\ \cos{3x} = 0\]

\[3x = \pm \arccos 0 + \pi n\]

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

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

\[x = \pm \frac{\pi}{6} + \frac{\text{πn}}{3}.\]

\[Имеет\ смысл\ при:\]

\[\sin{4x} \neq 0\]

\[4x \neq \arcsin 0 + \pi n \neq \pi n\]

\[x \neq \frac{\text{πn}}{4}.\]

\[\sin{3x} \neq 0\]

\[3x \neq \arcsin 0 + \pi n \neq \pi n\]

\[x \neq \frac{\text{πn}}{3}.\]

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

\[2)\ tg\ 2x + ctg\ x = 8\cos^{2}x\]

\[\frac{\sin{2x}}{\cos{2x}} + \frac{\cos x}{\sin x} - 8\cos^{2}x = 0\]

\[\frac{\sin{2x} \bullet \sin x + \cos x \bullet \cos{2x}}{\cos{2x} \bullet \sin x} - 8\cos^{2}x = 0\]

\[\frac{\cos(2x - x)}{\cos{2x} \bullet \sin x} - 8\cos^{2}x = 0\]

\[\cos x \bullet \left( \frac{1}{\cos{2x} \bullet \sin x} - 8\cos x \right) = 0\]

\[\cos x \bullet \frac{1 - 8\sin x \bullet \cos x \bullet \cos{2x}}{\cos{2x} \bullet \sin x} = 0\]

\[\cos x \bullet \frac{1 - 4\sin{2x} \bullet \cos{2x}}{\cos{2x} \bullet \sin x} = 0\]

\[\cos x \bullet \frac{1 - 2\sin{4x}}{\cos{2x} \bullet \sin x} = 0\]

\[1)\ \cos x = 0\]

\[x = \arccos 0 + \pi n = \frac{\pi}{2} + \pi n.\]

\[2)\ 1 - 2\sin{4x} = 0\]

\[2\sin{4x} = 1\]

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

\[4x = ( - 1)^{n} \bullet \arcsin\frac{1}{2} + \pi n\]

\[4x = ( - 1)^{n} \bullet \frac{\pi}{6} + \pi n\]

\[x = \frac{1}{4} \bullet \left( ( - 1)^{n} \bullet \frac{\pi}{6} + \pi n \right)\]

\[x = ( - 1)^{n} \bullet \frac{\pi}{24} + \frac{\text{πn}}{4}.\]

\[Имеет\ смысл\ при:\]

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

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

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

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

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

\[\sin x \neq 0\]

\[x \neq \arcsin 0 + \pi n \neq \pi n.\]

\[Ответ:\ \ \frac{\pi}{2} + \pi n;\ \ \]

\[( - 1)^{n} \bullet \frac{\pi}{24} + \frac{\text{πn}}{4}.\]