Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 673

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

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

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

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

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

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

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

\[\cos x = 0\]

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

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

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

\[4\sin^{2}x = 1\]

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

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

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

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

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

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

\[Ответ:\ \ \frac{\pi}{2} + \pi n;\ \ \pm \frac{\pi}{6} + \pi n.\]

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

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

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

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

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

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

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

\[D = 25 - 16 = 9\]

\[y_{1} = \frac{5 - 3}{2 \bullet 4} = \frac{2}{8} = \frac{1}{4};\text{\ \ }\]

\[y_{2} = \frac{5 + 3}{2 \bullet 4} = 1.\]

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

\[\cos x = \pm 1\]

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

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

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

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

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

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

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

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

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

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

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

\[3)\sin{4x} = 6\cos^{2}{2x} - 4\]

\[2\sin{2x} \bullet \cos{2x} =\]

\[= 6\cos^{2}{2x} - 4\left( \cos^{2}{2x} + \sin^{2}{2x} \right)\]

\[2\sin{2x} \bullet \cos{2x} =\]

\[= 6\cos^{2}{2x} - 4\cos^{2}{2x} - 4\sin^{2}{2x}\]

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

\[y = tg\ 2x:\]

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

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

\[D = 1^{2} + 4 \bullet 2 = 1 + 8 = 9\]

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

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

\[1)\ tg\ 2x = - 1\]

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

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

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

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

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

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

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

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

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

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

\[4)\ 2\cos^{2}{3x} + \sin{5x} = 1\]

\[2\cos^{2}{3x} + \sin{5x} - 1 = 0\]

\[\cos^{2}{3x} - \sin^{2}{3x} + \cos\left( \frac{\pi}{2} - 5x \right) = 0\]

\[\cos{6x} + \cos\left( \frac{\pi}{2} - 5x \right) = 0\]

\[2 \bullet \cos\frac{6x + \frac{\pi}{2} - 5x}{2} \bullet \cos\frac{6x - \frac{\pi}{2} + 5x}{2} = 0\]

\[\cos\left( \frac{x}{2} + \frac{\pi}{4} \right) \bullet \cos\left( \frac{11x}{2} - \frac{\pi}{4} \right) = 0\]

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

\[\frac{x}{2} + \frac{\pi}{4} = \arccos 0 + \pi n\]

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

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

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

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

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

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

\[2)\ \cos\left( \frac{11x}{2} - \frac{\pi}{4} \right) = 0\]

\[\frac{11x}{2} - \frac{\pi}{4} = \arccos 0 + \pi n\]

\[\frac{11x}{2} - \frac{\pi}{4} = \frac{\pi}{2} + \pi n\]

\[\frac{11x}{2} = \frac{\pi}{2} + \frac{\pi}{4} + \pi n\]

\[\frac{11x}{2} = \frac{2\pi}{4} + \frac{\pi}{4} + \pi n\]

\[\frac{11x}{2} = \frac{3\pi}{4} + \pi n\]

\[x = \frac{2}{11} \bullet \left( \frac{3\pi}{4} + \pi n \right)\]

\[x = \frac{6\pi}{44} + \frac{2\pi n}{11}\]

\[x = \frac{3\pi}{22} + \frac{2\pi n}{11}.\]

\[Ответ:\ \ \frac{\pi}{2} + 2\pi n;\ \ \frac{3\pi}{22} + \frac{2\pi n}{11}.\]