ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1198

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

\[= 1 + 3\cos x\]

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

\[- 1 - 3\cos x = 0\]

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

\[- 3\cos x + 3 = 0\]

\[4\cos^{2}x\left( \cos x - 1 \right) -\]

\[- 3 \cdot \left( \cos x - 1 \right) = 0\]

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

\[1)\cos x = 1\]

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

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

\[2 \cdot \left( 1 + \cos x \right) - 3 = 0\]

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

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

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

\[2)\ 8\sin^{3}x + 4\cos^{2}x =\]

\[= 1 + 6\sin x\]

\[8\sin^{3}x + 4 - 4\sin^{2}x - 1 -\]

\[- 6\sin x = 0\]

\[4\sin^{2}x\left( 2\sin x - 1 \right) -\]

\[- 3 \cdot \left( 2\sin x - 1 \right) = 0\]

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

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

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

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

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

\[2 \cdot \left( 1 - \cos x \right) - 3 = 0\]

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

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

\[Ответ:\ ( - 1)^{n} \cdot \frac{\pi}{6} + \pi k;\ \]

\[\pm \frac{2\pi}{3} + 2\pi k.\ \]

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

\[1 + tg^{2}x - 3 - tgx = 0\]

\[tg^{2}x - tgx - 2 = 0\]

\[Пусть\ tgx = y:\]

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

\[y_{1} + y_{2} = 1;\ \ \ y_{1} \cdot y_{2} = - 2\]

\[y_{1} = 2;\ \ \ y_{2} = - 1.\]

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

\[x = arctg\ 2 + \pi k.\]

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

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

\[Ответ:\ arctg\ 2 + \pi k;\ \ - \frac{\pi}{4} + \pi k.\]

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

\[2 \cdot \left( 1 + ctg^{2}2x \right) = ctg\ 2x + 5\]

\[2 + 2ctg^{2}2x - ctg\ 2x - 5 = 0\]

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

\[Пусть\ ctg\ 2x = y:\]

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

\[D = 1 + 24 = 25\]

\[y_{1} = \frac{1 + 5}{4} = \frac{3}{2};\ \ \]

\[y_{2} = \frac{1 - 5}{4} = - 1.\]

\[1)\ ctg\ 2x = \frac{3}{2}\]

\[2x = arctg\frac{3}{2} + \pi k\]

\[x = \frac{1}{2}\text{arctg}\frac{3}{2} + \frac{\text{πk}}{2}.\]

\[2)\ ctg\ 2x = - 1\]

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

\[x = \frac{3\pi}{8} + \frac{\text{πk}}{2}.\]

\[Ответ:\ \ \frac{1}{2}\text{arctg}\frac{3}{2} + \frac{\text{πk}}{2};\ \]

\[\ \frac{3\pi}{8} + \frac{\text{πk}}{2}\text{.\ }\]