Решебник по алгебре 11 класс Никольский Параграф 10. Равносильность уравнений на множествах Задание 48

\[\boxed{\mathbf{48.}}\]

\[\textbf{а)}\sin{2x} + 2\sin x - \sqrt{3}\cos x =\]

\[= \sqrt{3}\]

\[x \in (0;\pi);\]

\[\left( \cos x + 1 \right)\left( 2\sin x - \sqrt{3} \right) = 0\]

\[1)\cos x + 1 = 0\]

\[\cos x = - 1\]

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

\[0 < \pi + 2\pi k < \pi\]

\[0 < 1 + 2k < 1\]

\[- 1 < 2k < 0\]

\[- \frac{1}{2} < k < 0.\]

\[нет\ решений.\]

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

\[2\sin x = \sqrt{3}\]

\[\sin x = \frac{\sqrt{3}}{2}\]

\[x_{1} = \frac{\pi}{3} + 2\pi k;\]

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

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

\[k = 0:\]

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

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

\[k = 0:\]

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

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

\[\textbf{б)}\sin{2x} - 2\sin x + \sqrt{3}\cos x =\]

\[= \sqrt{3}\]

\[x \in \left\lbrack - \frac{\pi}{2};\frac{3\pi}{2} \right\rbrack;\]

\[\left( \cos x - 1 \right)\left( 2\sin x + \sqrt{3} \right) = 0\]

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

\[\cos x = 1\]

\[x = 2\pi k.\]

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

\[- \frac{1}{2} \leq 2k \leq \frac{3}{2}\]

\[- \frac{1}{4} \leq k \leq \frac{3}{4}\]

\[k = 0:\]

\[x = 0.\]

\[2)\ 2\sin x + \sqrt{3} = 0\]

\[2\sin x = - \sqrt{3}\]

\[\sin x = - \frac{\sqrt{3}}{2}\]

\[x_{1} = - \frac{\pi}{3} + 2\pi k;\]

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

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

\[k = 0:\]

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

\[- \frac{\pi}{2} \leq \frac{4\pi}{3} + 2\pi k \leq \frac{3\pi}{2}\]

\[k = 0:\]

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

\[Ответ:x = 0;\ \ x = \frac{\pi}{3};x = \frac{4\pi}{3}.\]