Решебник по алгебре 11 класс Никольский Параграф 10. Равносильность уравнений на множествах Задание 49

\[\boxed{\mathbf{49.}}\]

\[\textbf{а)}\ sin^{2}x - \sin x - 3cos^{2}x = 0\]

\[x \in \left( - \frac{\pi}{2};\frac{\pi}{2} \right).\]

\[\sin^{2}x - 2\sin x\cos x - 3\cos^{2}x =\]

\[= 0\ \ |\ :cos^{2}x\]

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

\[tgx = t:\]

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

\[D_{1} = 1 + 3 = 4\]

\[t_{1} = 1 + 2 = 3;\]

\[t_{2} = 1 - 2 = - 1.\]

\[1)\ tgx = 3\]

\[x = arctg\ 3 + \pi k\]

\[- \frac{\pi}{2} < arctg\ 3 + \pi k < \frac{\pi}{2}\]

\[k = 0:\]

\[- \frac{\pi}{2} < arctg\ 3 < \frac{\pi}{2}\]

\[x = arctg\ 3.\]

\[2)\ tgx = - 1\]

\[x = arctg\ ( - 1) + \pi k\]

\[x = - arctg\ 1 + \pi k\]

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

\[- \frac{\pi}{2} < - \frac{\pi}{4} + \pi k < \frac{\pi}{2}\]

\[k = 0:\]

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

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

\[\textbf{б)}\ 5sin^{2}x - 2\sin{2x} - cos^{2}x = 0\]

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

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

\[= 0\ \ \ |\ :cos^{2}x\]

\[5tg^{2}x - 4tg\ x - 1 = 0\]

\[tgx = t:\]

\[5t^{2} - 4t - 1 = 0\]

\[D_{1} = 4 + 5 = 9\]

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

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

\[1)\ tgx = 1\]

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

\[- \frac{\pi}{2} < \frac{\pi}{4} + 2\pi k < \frac{\pi}{2}\]

\[k = 0:\]

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

\[2)\ tgx = - \frac{1}{5}\]

\[x = - arctg\frac{1}{5} + \pi k\]

\[k = 0:\]

\[x = - arctg\frac{1}{5}.\]

\[Ответ:x = \frac{\pi}{4};x = - arctg\frac{1}{5}.\]