Решебник по алгебре 11 класс Никольский Параграф 10. Равносильность уравнений на множествах Задание 53

\[\boxed{\mathbf{53.}}\]

\[\textbf{а)}\ 3tg^{2}\left( \pi x - \frac{\pi}{8} \right) = 1\]

\[x \in \left( \frac{3}{2};3 \right);\]

\[tg^{2}\left( \pi x - \frac{\pi}{8} \right) = \frac{1}{3}\]

\[1)\ tg\left( \pi x - \frac{\pi}{8} \right) = \frac{1}{\sqrt{3}}\]

\[\frac{3}{2} < \frac{7}{24} + k < 3\]

\[k = 2:\]

\[x = 2\frac{7}{24}.\]

\[2)\ tg\ \left( \pi x - \frac{\pi}{8} \right) = - \frac{1}{\sqrt{3}}\]

\[\frac{3}{2} < - \frac{1}{24} + k < 3\]

\[k = 2:\]

\[x = - \frac{1}{24} + 2 = 1\frac{23}{24}.\]

\[k = 3:\]

\[x = - \frac{1}{24} + 3 = 2\frac{23}{24}.\]

\[Ответ:x = 2\frac{7}{24};x = 1\frac{23}{24};\]

\[x = 2\frac{23}{24}.\]

\[\textbf{б)}3\cos{2x} - 5\cos x = 1\]

\[x \in \lbrack 0;2\pi\rbrack;\]

\[3\left( 2cos^{2}x - 1 \right) - 5\cos x = 1\]

\[6cos^{2} - 3 - 5\cos x - 1 = 0\]

\[6\cos^{2}x - 5\cos x - 4 = 0\]

\[\cos x = t:\]

\[6t^{2} - 5t - 4 = 0\]

\[D = 25 + 96 = 121\]

\[t_{1} = \frac{5 + 11}{12} = \frac{16}{12} = \frac{4}{3};\]

\[t_{2} = \frac{5 - 11}{12} = - \frac{6}{12} = - \frac{1}{2}.\]

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

\[\cos x > 1\]

\[корней\ нет.\]

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

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

\[0 \leq \frac{2\pi}{3} + 2\pi k \leq 2\pi\]

\[k = 0:\]

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

\[0 \leq - \frac{2\pi}{3} + 2\pi k \leq 2\pi\]

\[k = 1:\]

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

\[Ответ:x = \frac{2\pi}{3};\ \ x = \frac{4\pi}{3}.\]

## Параграф 11. Равносильность неравенств на множествах