Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 1260

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\[1)\ \frac{\left( \sqrt{3} + 1 \right)\sin{3x} + \sin{5x}}{\left| \sin x \right|} =\]

\[= \sqrt{3}\]

\[1)\sin x > 0:\]

\[16\cos^{4}x + \left( 4\sqrt{3} - 8 \right)\cos^{2}x -\]

\[- \sqrt{3} = \sqrt{3}\]

\[Пусть\ t = \cos^{2}x:\]

\[16t^{2} + \left( 4\sqrt{3} - 8 \right)t - 2\sqrt{3} = 0\]

\[t_{1} = \frac{1}{2};\ \ \ \ t_{2} = - \frac{\sqrt{3}}{4}.\]

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

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

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

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

\[Подходят\ корни:\]

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

\[2)\sin x < 0:\]

\[16\cos^{4}x +\]

\[+ \left( 4\sqrt{3} - 8 \right)\cos^{2}x = 0\]

\[\cos x = 0\]

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

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

\[\cos x = \pm \left( \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} \right)\]

\[x = \pm \arccos{\left( \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} \right) + 2\pi n}\]

\[x =\]

\[= \pm \left( \pi - \arccos\left( \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} \right) \right) +\]

\[+ 2\pi n.\]

\[Нам\ подходят:\]

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

\[x = - \arccos\left( \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} \right) + 2\pi n;\]

\[x = - \pi + \arccos\left( \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} \right) +\]

\[+ 2\pi n.\]

\[2)\ \frac{\left( \sqrt{3} + 1 \right)\cos{3x} - \cos{5x}}{\left| \cos x \right|} =\]

\[= \sqrt{3}\]

\[1)\cos x > 0:\]

\[- 16\cos^{4}x +\]

\[+ \left( 4\sqrt{3} + 24 \right)\cos^{2}x - 3\sqrt{3} -\]

\[- 8 = \sqrt{3}\]

\[Пусть\cos^{2}x = t:\]

\[- 16t^{2} + \left( 4\sqrt{3} + 24 \right)t -\]

\[- 4\sqrt{3} - 8 = 0\]

\[Корни:\ \ t_{1} = 1;\ \ t_{2} = \frac{\sqrt{3}}{4} + \frac{1}{2}.\]

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

\[\cos x = \pm 1.\]

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

\[\cos x = \pm \left( \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} \right).\]

\[Нам\ подходят\ корни:\]

\[\cos x = 1\]

\[x = 2\pi n.\]

\[\cos x = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}\]

\[x = \pm \arccos\left( \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} \right).\]

\[2)\cos x < 0:\]

\[- 16\cos^{4}x +\]

\[+ \left( 4\sqrt{3} + 24 \right)\cos^{2}x -\]

\[- 3\sqrt{3} - 8 = - \sqrt{3}\]

\[Пусть\cos^{2}x = t:\]

\[- 16t^{2} + \left( 4\sqrt{3} + 24 \right)t -\]

\[- 2\sqrt{3} - 8 = 0\]

\[Корни:\ \ t_{1} = \frac{1}{2};\ \ t_{2} = \frac{\sqrt{3}}{4} + 1.\]

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

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

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

\[Ответ:\ \pm \arccos\left( \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} \right);\ \ \]

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

\[3)\ \frac{2\sin{3x}}{\sin x} = \frac{\left| \cos{6x} \right|}{\cos{2x}}\]

\[1)\cos{6x} \geq 0:\]

\[8\cos^{2}x - 2 = 16\cos^{4}x -\]

\[- 16\cos^{2}x + 1\]

\[16\cos^{4}x - 24\cos^{2}x + 3 = 0\]

\[Пусть\cos^{2}x = t:\]

\[16t^{2} - 24t + 3 = 0\]

\[D_{1} = 144 - 48 = 96\]

\[t_{1} = \frac{12 + 4\sqrt{6}}{16} = \frac{3 + \sqrt{6}}{4};\ \ \ \ \]

\[t_{2} = \frac{12 - 4\sqrt{6}}{16} = \frac{3 - \sqrt{6}}{4}.\]

\[\cos^{2}x = \frac{3 - \sqrt{6}}{4}\]

\[\cos x = \pm \frac{1}{2}\sqrt{3 - \sqrt{6}}\]

\[x = \pm \arccos\left( \pm \frac{1}{2}\sqrt{3 - \sqrt{6}} \right) +\]

\[+ 2\pi n.\]

\[2)\cos{6x} < 0\]

\[8\cos^{2}x - 2 =\]

\[= - (16\cos^{4}x - 16\cos^{2}x + 1)\]

\[16\cos^{4}x - 8\cos^{2}x - 1 = 0\]

\[Пусть\cos^{2}x = t:\]

\[16t^{2} - 8t - 1 = 0\]

\[D_{1} = 16 + 16 = 32\]

\[t_{1,2} = \frac{4 \pm 4\sqrt{2}}{16} = \frac{1 \pm \sqrt{2}}{4}.\]

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

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

\[+ 2\pi n.\]

\[Ответ:\ \]

\[\pm \arccos\left( \pm \frac{1}{2}\sqrt{3 - \sqrt{6}} \right) + 2\pi n;\ \ \]

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

\[4)\ \frac{\sin{6x}}{\left| \sin{4x} \right|} = \frac{\cos{3x}}{\cos x}\]

\[1)\sin{4x} > 0:\]

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

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

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

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

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

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

\[Нам\ подходит\ корень:\]

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

\[2)\sin{4x} < 0:\]

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

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

\[Пусть\cos{2x} = t:\]

\[2t - \frac{1}{t} = - 2t + 1\]

\[4t^{2} - 2t - 1 = 0\]

\[t_{1} = \frac{1}{2};\ \ \ t_{2} = - \frac{1}{4}.\]

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

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

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

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

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

\[Нам\ подходят\ корни:\]

\[x = - \frac{\pi}{6} + 2\pi n;\ \ \]

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

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

\[\text{\ \ }\frac{1}{2}\arccos\left( - \frac{1}{4} \right) + \pi n.\ \ \ \]