Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1385

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\[1)\ tg\ 2x = 3\ tg\ x\]

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

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

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

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

\[\sin x = 0\]

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

\[3 - 4\cos^{2}x = 0\]

\[4\cos^{2}x = 3\]

\[\cos^{2}x = \frac{3}{4}\]

\[\cos x = \pm \frac{\sqrt{3}}{2}\]

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

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

\[Имеет\ смысл\ при:\]

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

\[x \neq \frac{1}{2} \bullet \left( \frac{\pi}{2} + \pi n \right) = \frac{\pi}{4} + \frac{\text{πn}}{2}.\]

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

\[2)\ ctg\ 2x = 2\ ctg\ x\]

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

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

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

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

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

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

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

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

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

\[корней\ нет.\]

\[Ответ:\ \ решений\ нет.\]

\[3)\ tg\left( x + \frac{\pi}{4} \right) + tg\left( x - \frac{\pi}{4} \right) = 2\]

\[\frac{tg\ x + tg\frac{\pi}{4}}{1 - tg\ x \bullet tg\frac{\pi}{4}} + \frac{tg\ x - tg\frac{\pi}{4}}{1 + tg\ x \bullet tg\frac{\pi}{4}} = 2\]

\[\frac{tg\ x + 1}{1 - tg\ x} + \frac{tg\ x - 1}{1 + tg\ x} = 2\]

\[\frac{(tg\ x + 1)^{2} - (1 - tg\ x)^{2}}{(1 - tg\ x)(1 + tg\ x)} = 2\]

\[\frac{4 \bullet tg\ x}{1 - tg^{2}\text{\ x}} = 2\]

\[2\ tg\ 2x = 2\]

\[tg\ 2x = 1\]

\[2x = arctg\ 1 + \pi n = \frac{\pi}{4} + \pi n\]

\[x = \frac{1}{2} \bullet \left( \frac{\pi}{4} + \pi n \right) = \frac{\pi}{8} + \frac{\text{πn}}{2}.\]

\[Ответ:\ \ \frac{\pi}{8} + \frac{\text{πn}}{2}.\]

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

\[\text{tg}(2x + 1) = \frac{1}{\text{ctg}(x + 1)}\]

\[\text{tg}(2x + 1) = tg(x + 1)\]

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

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

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

\[\sin x = 0\]

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

\[Ответ:\ \ \pi n.\]