Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 643

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\[1)\ \sqrt{5\cos x - \cos{2x}} = - 2\sin x\]

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

\[5\cos x - \left( \cos^{2}x - \sin^{2}x \right) - 4\sin^{2}x = 0\]

\[5\cos x - \cos^{2}x + \sin^{2}x - 4\left( 1 - \cos^{2}x \right) = 0\]

\[5\cos x - \cos^{2}x + \sin^{2}x - 4 + 4\cos^{2}x = 0\]

\[5\cos x + 3\cos^{2}x + \sin^{2}x - 4 = 0\]

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

\[2\cos^{2}x + 5\cos x - 3 = 0\]

\[y = \cos x:\]

\[2y^{2} + 5y - 3 = 0\]

\[D = 25 + 24 = 49\]

\[y_{1} = \frac{- 5 - 7}{2 \bullet 2} = - 3;\]

\[y_{2} = \frac{- 5 + 7}{2 \bullet 2} = \frac{1}{2}.\]

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

\[корней\ нет.\]

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

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

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

\[Проверка:\]

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

\[= 2 \bullet \frac{\sqrt{3}}{2} = \sqrt{3};\]

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

\[= - 2 \bullet \frac{\sqrt{3}}{2} = - \sqrt{3} - не\ подходит.\]

\[Ответ:\ - \frac{\pi}{3} + 2\pi n.\]

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

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

\[\sqrt{\cos{2x} \bullet \cos x} = - \cos x\]

\[(\cos^{2}x - \sin^{2}x) \bullet \cos x = \cos^{2}x\]

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

\[\left( 2\cos^{2}x - 1 \right) \bullet \cos x - \cos^{2}x = 0\]

\[\cos x \bullet \left( 2\cos^{2}x - \cos x - 1 \right) = 0\]

\[y = \cos x:\]

\[2y^{2} - y - 1 = 0\]

\[D = 1 + 8 = 9\]

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

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

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

\[x = \arccos 0 + \pi n\]

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

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

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

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

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

\[3)\ \cos x = 1\]

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

\[Проверка:\]

\[- \sqrt{2}\cos\left( \frac{\pi}{2} + \pi n \right) = - \sqrt{2} \bullet 0 = 0\]

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

\[= - \sqrt{2}\cos\left( \frac{2\pi}{3} \right) =\]

\[= - \sqrt{2}\cos\left( \pi - \frac{\pi}{3} \right) =\]

\[= \sqrt{2}\cos\frac{\pi}{3} = \frac{\sqrt{2}}{2};\]

\[- \sqrt{2}\cos\left( \text{πn} \right) = - \sqrt{2} \bullet 1 =\]

\[= - \sqrt{2} - не\ подходит.\]

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