Решебник по алгебре 11 класс Никольский Параграф 10. Равносильность уравнений на множествах Задание 21

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\[\textbf{а)}\ 3\cos x + 4\sin x = 0\ \ | \cdot \frac{1}{\cos x}\]

\[\cos x \neq 0;\]

\[3 + \frac{4\sin x}{\cos x} = 0\]

\[3 + 4tg\ x = 0\]

\[tgx = - \frac{3}{4}\]

\[x = - arctg\ \left( \frac{3}{4} \right) + \pi n.\]

\[Ответ:x = - arctg\ \left( \frac{3}{4} \right) + \pi n.\]

\[\textbf{б)}\ 2\sin x - \cos x = 0\ \ \ \ | \cdot \frac{1}{\cos x}\]

\[\cos x \neq 0;\]

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

\[2tgx = 1\]

\[tgx = \frac{1}{2}\]

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

\[Ответ:x = arctg\ \left( \frac{1}{2} \right) + \pi n.\]

\[\text{co}s^{2}x \neq 0;\]

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

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

\[tgx = t:\]

\[2t^{2} - 3t + 1 = 0\]

\[D = 9 - 8 = 1\]

\[t_{1} = \frac{3 + 1}{4} = 1;\]

\[t_{2} = \frac{3 - 1}{4} = \frac{1}{2}.\]

\[tgx = 1\]

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

\[tgx = \frac{1}{2}\]

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

\[Ответ:x = \frac{\pi}{4} + \pi k;\]

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

\[\text{co}s^{2}x \neq 0;\]

\[\frac{3sin^{2}x}{\text{co}s^{2}x} - \frac{5\sin x}{\cos x} + 2 = 0\]

\[3tg^{2}x - 5tgx + 2 = 0\]

\[tgx = t:\]

\[3t^{2} - 5t + 2 = 0\]

\[D = 25 - 24 = 1\]

\[t_{1} = \frac{5 + 1}{6} = 1;\]

\[t_{2} = \frac{5 - 1}{6} = \frac{2}{3}.\]

\[tgx = 1\]

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

\[tgx = \frac{2}{3}\]

\[x = arctg\ \left( \frac{2}{3} \right) + \pi n.\]

\[Ответ:\ x = \frac{\pi}{4} + \pi k;\]

\[x = arctg\ \left( \frac{2}{3} \right) + \pi n.\]