Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1563

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\[1)\ tg\ x + ctg\ x = 2\ ctg\ 4x\]

\[tg\ x + ctg\ x = 2 \bullet \frac{\text{ct}g^{2}\ 2x - 1}{2 \bullet ctg\ 2x}\]

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

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

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

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

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

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

\[\frac{2\cos^{2}{2x} - 2\cos{2x} - 1}{0,5\sin{4x}} = 0\]

\[y = \cos{2x}:\]

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

\[D = 4 + 8 = 12 = 4 \bullet 3\]

\[y = \frac{2 \pm \sqrt{12}}{2 \bullet 2} = \frac{2 \pm 2\sqrt{3}}{2 \bullet 2} =\]

\[= \frac{1 \pm \sqrt{3}}{2}.\]

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

\[корней\ нет.\]

\[2)\ \cos{2x} = \frac{1 - \sqrt{3}}{2}\]

\[2x = \pm \arccos\frac{1 - \sqrt{3}}{2} + 2\pi n\]

\[x = \pm \frac{1}{2}\arccos\frac{1 - \sqrt{3}}{2} + \pi n.\]

\[Имеет\ смысл\ при:\]

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

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

\[x \neq \frac{\text{πn}}{4}.\]

\[Ответ:\ \ \pm \frac{1}{2}\arccos\frac{1 - \sqrt{3}}{2} + \pi n.\]

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

\[\frac{\sin{4x}}{\sin x \bullet \cos\frac{\pi}{4} - \sin\frac{\pi}{4} \bullet \cos x} =\]

\[= \sqrt{2}\left( \sin x + \cos x \right)\]

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

\[= \sqrt{2}\left( \sin x + \cos x \right)\]

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

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

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

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

\[1)\ \cos{2x} = 0\]

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

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

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

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

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

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

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

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

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

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

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

\[Имеет\ смысл\ при:\]

\[\sin x - \cos x \neq 0\ \ \ \ \ |\ :\cos x\]

\[tg\ x - 1 \neq 0\]

\[tg\ x \neq 1\]

\[x \neq arctg\ 1 + \pi n\]

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

\[Ответ:\ \ \]

\[- \frac{\pi}{4} + \pi n;\ \ ( - 1)^{n + 1} \bullet \frac{\pi}{12} + \frac{\text{πn}}{2}\text{.\ \ }\]