ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1266

\[\boxed{\mathbf{1266}\mathbf{.}}\]

\[1)\sin\frac{3x}{2} + \cos\frac{3x}{2} > 0\]

\[\sqrt{2}\sin\left( \frac{3x}{2} + \frac{\pi}{4} \right) > 0\]

\[2\pi n < \frac{3x}{2} + \frac{\pi}{4} < \pi + 2\pi n\]

\[- \frac{\pi}{4} + 2\pi n < \frac{3x}{2} < \frac{3\pi}{4} + 2\pi n\]

\[- \frac{\pi}{6} + \frac{4\pi n}{3} < x < \frac{\pi}{2} + \frac{4\pi n}{3}.\]

\[2)\cos{2x} + \cos x \geq 0\]

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

\[Пусть\cos x = y:\]

\[2y^{2} + y - 1 \geq 0\]

\[D = 1 + 8 = 9\]

\[y_{1} = \frac{- 1 + 3}{4} = \frac{1}{2};\ \ \]

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

\[(y + 1)\left( y - \frac{1}{2} \right) \geq 0\]

\[y \leq - 1;\ \ y \geq \frac{1}{2}.\]

\[\cos x = - 1\]

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

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

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

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