Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 879

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

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

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

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

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

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

\[D = 3^{2} - 4 \bullet 2 = 9 - 8 = 1\]

\[\cos x_{1} = \frac{3 - 1}{2} = 1;\]

\[\cos x_{2} = \frac{3 + 1}{2} = 2;\]

\[\cos x = 0\]

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

\[\cos x = 1\]

\[x = 2\pi n.\]

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

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

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

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

\[\sin{2x} + \sin x \bullet \sin{2x} \bullet \sin{3x} = 0\]

\[\sin{2x} \bullet \left( 1 + \sin x \bullet \sin{3x} \right) = 0\]

\[\sin{2x} = 0\]

\[2x = \pi n\]

\[x = \frac{\text{πn}}{2}.\]

\[Область\ определения:\]

\[x \neq \pi n;\ \ \ x \neq \frac{\text{πn}}{3}.\]

\[Ответ:\ \ \frac{\pi}{2} + \pi n.\]