Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 674

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

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

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

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

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

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

\[4y^{2} + 4y - 3 = 0\]

\[D = 16 + 48 = 64\]

\[y_{1} = \frac{- 4 - 8}{2 \bullet 4} = - \frac{12}{8} = - \frac{3}{2};\]

\[y_{2} = \frac{- 4 + 8}{2 \bullet 4} = \frac{4}{8} = \frac{1}{2}.\]

\[1)\ \cos{2x} = - \frac{3}{2}\]

\[корней\ нет.\]

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

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

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

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

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

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

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

\[\sin{3x} + \sin x = 4\sin x\]

\[2 \bullet \sin\frac{3x + x}{2} \bullet \cos\frac{3x - x}{2} - 4\sin x = 0\]

\[2 \bullet \sin{2x} \bullet \cos x - 4\sin x = 0\]

\[4\sin x \bullet \cos x \bullet \cos x - 4\sin x = 0\]

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

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

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

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

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

\[\cos x = \pm 1\]

\[x_{1} = \pi - \arccos 1 + \pi n =\]

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

\[x_{2} = \arccos 1 + \pi n = 2\pi n.\]

\[Ответ:\ \ \pi n.\]

\[3)\ 3\cos{2x} - 7\sin x = 4\]

\[3\left( \cos^{2}x - \sin^{2}x \right) - 7\sin x - 4 = 0\]

\[3\left( 1 - \sin^{2}x - \sin^{2}x \right) - 7\sin x - 4 = 0\]

\[3 - 6\sin^{2}x - 7\sin x - 4 = 0\]

\[6\sin^{2}x + 7\sin x + 1 = 0\]

\[y = \sin x:\]

\[6y^{2} + 7y + 1 = 0\]

\[D = 49 - 24 = 25\]

\[y_{1} = \frac{- 7 - 5}{2 \bullet 6} = - 1;\]

\[y_{2} = \frac{- 7 + 5}{2 \bullet 6} = \frac{2}{12} = \frac{1}{6}.\]

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

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

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

\[2)\ \sin x = \frac{1}{6}\]

\[x = ( - 1)^{n} \bullet \arcsin\frac{1}{6} + \pi n.\]

\[Ответ:\ - \frac{\pi}{2} + 2\pi n;\ \ \]

\[( - 1)^{n} \bullet \arcsin\frac{1}{6} + \pi n.\]

\[4)\ 1 + \cos x + \cos{2x} = 0\]

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

\[\cos x \bullet \left( 2\cos x + 1 \right) = 0\]

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

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

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

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

\[2\cos x = - 1\]

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

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

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

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

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

\[5)\ 5\sin{2x} + 4\cos^{3}x - 8\cos x = 0\]

\[10\sin x \bullet \cos x + 4\cos^{3}x - 8\cos x = 0\]

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

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

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

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

\[y = \sin x:\]

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

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

\[D = 25 - 16 = 9\]

\[y_{1} = \frac{5 - 3}{2 \bullet 2} = \frac{2}{4} = \frac{1}{2};\]

\[y_{2} = \frac{5 + 3}{2 \bullet 2} = 2.\]

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

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

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

\[x = ( - 1)^{n} \bullet \arcsin\frac{1}{2} + \pi n\]

\[x = ( - 1)^{n} \bullet \frac{\pi}{6} + \pi n.\]

\[3)\ \sin x = 2\]

\[корней\ нет.\]

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

\[\text{\ \ }( - 1)^{n} \bullet \frac{\pi}{6} + \pi n.\]