ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 576

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

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

\[\cos{4x} = 1\]

\[4x = \arccos 1 + 2\pi n =\]

\[= 0 + 2\pi n = 2\pi n\]

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

\[x = \frac{\text{πn}}{2}\]

\[Ответ:\ \ x = \frac{\text{πn}}{2}.\]

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

\[\cos^{2}x = \frac{3}{4}\]

\[\cos x = \pm \frac{\sqrt{3}}{2}\]

\[x_{1} = \pm \left( \pi - \arccos\frac{\sqrt{3}}{2} \right) + 2\pi n =\]

\[= \pm \left( \pi - \frac{\pi}{6} \right) + 2\pi n =\]

\[= \pm \frac{5\pi}{6} + 2\pi n\]

\[x_{2} = \pm \arccos\frac{\sqrt{3}}{2} + 2\pi n =\]

\[= \pm \frac{\pi}{6} + 2\pi n\]

\[Ответ:\ x = \pm \frac{\pi}{6} + \pi n.\]

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

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

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

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

\[2x = \pm \arccos\frac{1}{2} + 2\pi n =\]

\[= \pm \frac{\pi}{3} + 2\pi n\]

\[x = \frac{1}{2} \bullet \left( \pm \frac{\pi}{3} + 2\pi n \right)\]

\[x = \pm \frac{\pi}{6} + \pi n\]

\[Ответ:\ x = \pm \frac{\pi}{6} + \pi n.\]

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

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

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

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

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

\[2x = \pm \arccos\frac{1}{\sqrt{2}} + 2\pi n =\]

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

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

\[x = \pm \frac{\pi}{8} + \pi n\]

\[Ответ:\ x = \pm \frac{\pi}{8} + \pi n.\]

\[5)\ \left( 1 + \cos x \right)\left( 3 - 2\cos x \right) = 0\]

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

\[\cos x = - 1\]

\[x = \left( \pi - \arccos 1 \right) + 2\pi n\]

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

\[2)\ 3 - 2\cos x = 0\]

\[2\cos x = 3\]

\[\cos x = 1,5\]

\[корней\ нет.\]

\[Ответ:\ x = \ \pi + 2\pi n.\]

\[6)\ \left( 1 - \cos x \right)\left( 4 + 3\cos x \right) = 0\]

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

\[\cos x = 1\]

\[x = \arccos 1 + 2\pi n\]

\[x = 0 + 2\pi n = 2\pi n.\]

\[2)\ 4 + 3\cos x = 0\]

\[3\cos x = - 4\]

\[\cos x = - \frac{4}{3}\]

\[корней\ нет.\]

\[Ответ:\ \ x = 2\pi n.\]

\[7)\ \left( 1 + 2\cos x \right)\left( 1 - 3\cos x \right) = 0\]

\[1)\ 1 + 2\cos x = 0\]

\[2\cos x = - 1\]

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

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

\[= \pm \left( \pi - \frac{\pi}{3} \right) + 2\pi n\]

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

\[2)\ 1 - 3\cos x = 0\]

\[3\cos x = 1\]

\[\cos x = \frac{1}{3}\]

\[x = \pm \arccos\frac{1}{3} + 2\pi n\]

\[Ответ:\ x = \pm \frac{2\pi}{3} + 2\pi n;\ \]

\[x = \pm \arccos\frac{1}{3} + 2\pi n.\]

\[8)\ \left( 1 - 2\cos x \right)\left( 2 + 3\cos x \right) = 0\ \]

\[1)\ 1 - 2\cos x = 0\]

\[2\cos x = 1\]

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

\[x = \pm \arccos\frac{1}{2} + 2\pi n\]

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

\[2)\ 2 + 3\cos x = 0\]

\[3\cos x = - 2\]

\[\cos x = - \frac{2}{3}\]

\[x = \pm \left( \pi - \arccos\frac{2}{3} \right) + 2\pi n\]

\[Ответ:\ x = \pm \frac{\pi}{3} + 2\pi n;\ \]

\[x = \pm \left( \pi - \arccos\frac{2}{3} \right) + 2\pi n.\]