Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 634

\[\boxed{\mathbf{634}\mathbf{.}}\]

\[2 + 6\ tg\ 2x + 4\ tg^{2}\ 2x = 0\]

\[y = tg\ 2x:\]

\[4y^{2} + 6y + 2 = 0\]

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

\[D = 9 - 8 = 1\]

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

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

\[tg\ 2x = - 1\]

\[2x = - arctg\ 1 + \pi n\]

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

\[x = \frac{1}{2} \bullet \left( - \frac{\pi}{4} + \pi n \right)\]

\[x = - \frac{\pi}{8} + \frac{\text{πn}}{2}.\]

\[tg\ 2x = - \frac{1}{2}\]

\[2x = - arctg\frac{1}{2} + \pi n\]

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

\[x = - \frac{1}{2}\text{arctg}\frac{1}{2} + \frac{\text{πn}}{2}.\]

\[Ответ:\ - \frac{\pi}{8} + \frac{\text{πn}}{2};\ \ \]

\[- \frac{1}{2}\text{arctg}\frac{1}{2} + \frac{\text{πn}}{2}.\]

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

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

\[y = tg\ x:\]

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

\[D = 1 - 12 = - 11 < 0\]

\[корней\ нет.\]

\[Ответ:\ \ корней\ нет.\]

\[3)\ 2\sin^{2}x + \frac{1}{4}\cos^{3}{2x} = 1\]

\[2\sin^{2}x - \left( \cos^{2}x + \sin^{2}x \right) + \frac{1}{4}\cos^{3}{2x} = 0\]

\[- \left( \cos^{2}x - \sin^{2}x \right) + \frac{1}{4}\cos^{3}{2x} = 0\]

\[- \cos{2x} + \frac{1}{4}\cos^{3}{2x} = 0\]

\[y = \cos{2x}:\]

\[\frac{1}{4}y^{3} - y = 0\]

\[y\left( \frac{1}{4}y^{2} - 1 \right) = 0\]

\[1)\ \cos{2x} = 0\]

\[2x = \arccos 0 + \pi n = \frac{\pi}{2} + \pi n\]

\[x = \frac{1}{2} \bullet \left( \frac{\pi}{2} + \pi n \right) = \frac{\pi}{4} + \frac{\text{πn}}{2}.\]

\[2)\ \frac{1}{4}\cos^{2}{2x} - 1 = 0\]

\[\frac{1}{4}\cos^{2}{2x} = 1\]

\[\cos^{2}{2x} = 4\]

\[\cos{2x} = \pm 2\]

\[корней\ нет.\]

\[Ответ:\ \ \frac{\pi}{4} + \frac{\text{πn}}{2}.\]

\[4)\sin^{2}{2x} + \cos^{2}{3x} = 1 + 4\sin x\]

\[\sin^{2}{2x} - \left( 1 - \cos^{2}{3x} \right) = 4\sin x\]

\[\sin^{2}{2x} - \sin^{2}{3x} = 4\sin x\]

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

\[- \sin x \bullet \sin{5x} - 4\sin x = 0\]

\[- \sin x\left( \sin{5x} + 4 \right) = 0\]

\[1)\ \sin x = 0\]

\[x = \arcsin 0 + \pi n = \pi n.\]

\[2)\ \sin{5x} + 4 = 0\]

\[\sin{5x} = - 4\]

\[корней\ нет.\]

\[Ответ:\ \ \pi n.\]