ГДЗ по алгебре 11 класс Никольский Параграф 8. Уравнения-следствия Задание 27

\[\boxed{\mathbf{27.}}\]

\[\textbf{а)}\ tg\ 3x = tg\ 5x\]

\[\frac{\sin{3x}}{\cos{3x}} = \frac{\sin{5x}}{\cos{5x}}\]

\[\frac{\sin{3x}\cos{5x} - \sin{5x}\cos{3x}}{\cos{3x}\cos{5x}} = 0\]

\[\frac{\sin( - 2x)}{\cos{3x}\cos{5x}} = 0\]

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

\[2x = \pi n\]

\[x = \frac{\text{πn}}{2}.\]

\[\cos\left( 3 \cdot \frac{\text{πn}}{2} \right)\cos\left( 5 \cdot \frac{\text{πn}}{2} \right) = 0\ \]

\[при\ n - нечетное\ число;\]

\[\cos\left( 3 \cdot \frac{\text{πn}}{2} \right)\cos\left( 5 \cdot \frac{\text{πn}}{2} \right) \neq 0\ \]

\[при\ n - четное\ число.\]

\[Ответ:\ x = \frac{\text{πn}}{2};n - четное\ \]

\[число.\]

\[\textbf{б)}\ cyg\ 3x = ctg\ 5x\ \]

\[\frac{\cos{3x}}{\sin{3x}} - \frac{\cos{5x}}{\sin{5x}} = 0\]

\[\frac{\cos{3x}\sin{5x} - \cos{5x}\sin{3x}}{\sin{3x}\sin{5x}} = 0\]

\[\frac{\sin{2x}}{\sin{3x}\sin{5x}} = 0\]

\[\sin{2x} = 0\]

\[x = \frac{\text{πn}}{2}.\]

\[При\ n - четное\ число:\]

\[решений\ нет.\]

\[При\ n - нечетное\ число:\]

\[ctg\ 3 \cdot \frac{\pi(2m + 1)}{2} =\]

\[= ctg\ 5 \cdot \frac{\pi(2m + 1)}{2}\]

\[\text{ctg\ }\left( 3\pi m + \frac{3\pi}{2} \right) =\]

\[= ctg\ \left( 5\pi m + \frac{5\pi}{2} \right)\]

\[\text{ctg}\frac{3\pi}{2} = ctg\frac{5\pi}{2}\]

\[0 = 0.\]

\[Ответ:x = \frac{n}{2};\ \ n - нечетное\]

\[\ число.\]

\[\textbf{в)}\sin x = \frac{\cos x}{\sin x}\]

\[\frac{\text{si}n^{2}x - \cos x}{\sin x} = 0;\ \sin x \neq 0;\]

\[x \neq \text{πn}\]

\[\text{si}n^{2}x - \cos x = 0\]

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

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

\[\cos x = t;\ - 1 \leq t \leq 1:\]

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

\[D = 1 + 4 = 5\]

\[t = \frac{- 1 \pm \sqrt{5}}{2}.\]

\[\cos x =\]

\[= \frac{- 1 - \sqrt{5}}{2} < - 1\ (не\ подходит);\]

\[\cos x = \frac{- 1 + \sqrt{5}}{2}\]

\[x = \pm \arccos\left( \frac{- 1 + \sqrt{5}}{2} \right) + 2\pi n.\]

\[Ответ:x =\]

\[= \pm \arccos\left( \frac{- 1 + \sqrt{5}}{2} \right) + 2\pi n.\ \]

\[\textbf{г)}\cos x = \frac{\sin x}{\cos x}\ \]

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

\[\ \cos x \neq 0;\ x \neq \frac{\pi}{2} + \pi n\]

\[\text{co}s^{2}x - \sin x = 0\]

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

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

\[\sin x = t;\ - 1 \leq t \leq 1:\]

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

\[D = 1 + 4 = 5\]

\[t = \frac{- 1 \pm \sqrt{5}}{2}.\]

\[\sin x =\]

\[= \frac{- 1 - \sqrt{5}}{2} < - 1\ (не\ подходит);\]

\[\sin x = \frac{- 1 + \sqrt{5}}{2}\]

\[x = ( - 1)^{n}\arcsin\left( \frac{- 1 + \sqrt{5}}{2} \right) + \pi n.\]

\[Ответ:\ x =\]

\[= ( - 1)^{n}\arcsin\left( \frac{- 1 + \sqrt{5}}{2} \right) + \pi n.\]