ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1384

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\[1)\ tg^{2}\ 3x - 4\sin^{2}{3x} = 0\]

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

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

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

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

\[x = \frac{\text{πn}}{3}.\]

\[1 - 4\cos^{2}{3x} = 0\]

\[4\cos^{2}{3x} = 1\]

\[\cos^{2}{3x} = \frac{1}{4}\]

\[\cos{3x} = \pm \frac{1}{2}\]

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

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

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

\[x = \pm \frac{\pi}{9} + \frac{\text{πn}}{3}.\]

\[Имеет\ смысл\ при:\]

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

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

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

\[x \neq \frac{1}{3} \bullet \left( \frac{\pi}{2} + \pi n \right) \neq \frac{\pi}{6} + \frac{\text{πn}}{3}.\]

\[Ответ:\ \ \frac{\text{πn}}{3};\ \ \pm \frac{\pi}{9} + \frac{\text{πn}}{3}.\]

\[2)\sin x \bullet tg\ x = \cos x + tg\ x\]

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

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

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

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

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

\[y = \sin x:\]

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

\[D = 1^{2} + 4 \bullet 2 = 1 + 8 = 9\]

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

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

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

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

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

\[2)\ \sin x = 1\]

\[x = \arcsin 1 + 2\pi n = \frac{\pi}{2} + 2\pi n.\]

\[Имеет\ смысл\ при:\]

\[\cos x \neq 0\]

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

\[Ответ:\ \ ( - 1)^{n + 1} \bullet \frac{\pi}{6} + \pi n.\]

\[3)\ ctg\ x \bullet \left( ctg\ x + \frac{1}{\sin x} \right) = 1\]

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

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

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

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

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

\[y = \cos x:\]

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

\[D = 1 + 8 = 9\]

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

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

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

\[x = \pi - \arccos 1 + 2\pi n\]

\[x = \pi + 2\pi n.\]

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

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

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

\[Имеет\ смысл\ при:\]

\[\sin x \neq 0\]

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

\[Ответ:\ \ \pm \frac{\pi}{3} + 2\pi n.\]

\[4)\ 4\ ctg^{2}\ x = 5 - \frac{9}{\sin x}\]

\[4 \bullet \frac{\cos^{2}x}{\sin^{2}x} = \frac{5\sin x - 9}{\sin x}\ \ \ | \bullet \sin^{2}x\]

\[4\cos^{2}x = 5\sin^{2}x - 9\sin x\]

\[4 - 4\sin^{2}x - 5\sin^{2}x + 9\sin x = 0\]

\[- 9\sin^{2}x + 9\sin x + 4 = 0\]

\[y = \sin x:\]

\[- 9y^{2} + 9y + 4 = 0\]

\[9y^{2} - 9y - 4 = 0\]

\[D = 81 + 144 = 225\]

\[y_{1} = \frac{9 - 15}{2 \bullet 9} = - \frac{6}{18} = - \frac{1}{3};\]

\[y_{2} = \frac{9 + 15}{2 \bullet 9} = \frac{24}{18}.\]

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

\[x = ( - 1)^{n + 1} \bullet \arcsin\frac{1}{3} + \pi n.\]

\[2)\ \sin x = \frac{24}{18}\]

\[корней\ нет.\]

\[Имеет\ смысл\ при:\]

\[\sin x \neq 0\]

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

\[Ответ:\ \ ( - 1)^{n + 1} \bullet \arcsin\frac{1}{3} + \pi n.\]