Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 1208

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\[1)\ 1 - \cos(\pi - x) +\]

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

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

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

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

\[Первое\ уравнение:\]

\[\cos\frac{x}{2} = 0\]

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

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

\[Второе\ уравнение:\]

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

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

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

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

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

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

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

\[= \pm \frac{4\pi}{3} + 4\pi n.\]

\[Ответ:\ \ \pi + 2\pi n;\ \ \pm \frac{4\pi}{3} + 4\pi n.\]

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

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

\[\sqrt{2} \bullet\]

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

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

\[\sqrt{2} \bullet \left( \frac{1}{\sqrt{2}} \bullet \cos x + \frac{1}{\sqrt{2}} \bullet \sin x \right) =\]

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

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

\[Первое\ уравнение:\]

\[\sin x + \cos x = 0\ \ \ \ \ |\ :\cos x\]

\[tg\ x + 1 = 0\]

\[tg\ x = - 1\]

\[x = - arctg\ 1 + \pi n = - \frac{\pi}{4} + \pi n.\]

\[Второе\ уравнение:\]

\[\sin x + \cos x = 1\ \ \ \ \ |\ :\sqrt{2}\]

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

\[\sin\frac{\pi}{4} \bullet \sin x + \cos\frac{\pi}{4} \bullet \cos x = \frac{\sqrt{2}}{2}\]

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

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

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

\[x_{1} = - \frac{\pi}{4} + \frac{\pi}{4} + 2\pi n = 2\pi n;\]

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

\[Ответ:\ - \frac{\pi}{4} + \pi n;\ \ 2\pi n;\ \ \]

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

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

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

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

\[\frac{x}{2} = \frac{\pi}{2} + \pi k\]

\[x = \pi + 2\pi k.\]

\[1 + \frac{1 - \text{tg}^{2}\frac{x}{2}}{1 + \text{tg}^{2}\frac{x}{2}} = \frac{1}{\text{tg}\frac{x}{2}}\]

\[Пусть\ tg\frac{x}{2} = y:\]

\[1 + \frac{1 - y^{2}}{1 + y²} = \frac{1}{y}\]

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

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

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

\[y = 1\]

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

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

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

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

\[4)\sin x + tg\frac{x}{2} = 0\]

\[\frac{2\ tg\frac{x}{2}}{1 + tg^{2}\frac{x}{2}} + tg\frac{x}{2} = 0\]

\[Пусть\ tg\frac{x}{2} = y:\]

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

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

\[3y + y^{3} = 0\]

\[y\left( 3 + y^{2} \right) = 0\]

\[y = 0;\ \ \ y^{2} = - 3 \rightarrow нет\ корней.\]

\[\text{tg}\frac{x}{2} = 0\]

\[\frac{x}{2} = \pi k\]

\[x = 2\pi k.\]

\[Ответ:2\pi k.\]