Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 625

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\[1)\sin x - \cos x = 1\ \ \ \ \ |\ \bullet \frac{\sqrt{2}}{2}\]

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

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

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

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

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

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

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

\[x_{1} = - \frac{3\pi}{4} + 2\pi n - \frac{\pi}{4} = - \pi + 2\pi n;\]

\[x_{2} = + \frac{3\pi}{4} + 2\pi n - \frac{\pi}{4} = \frac{\pi}{2} + 2\pi n.\]

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

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

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

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

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

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

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

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

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

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

\[3)\ \sqrt{3}\sin x + \cos x = 2\ \ \ \ \ |\ :2\]

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

\[\cos\frac{\pi}{6} \bullet \sin x + \sin\frac{\pi}{6} \bullet \cos x = 1\]

\[\sin\left( \frac{\pi}{6} + x \right) = 1\]

\[x + \frac{\pi}{6} = \arcsin 1 + 2\pi n\]

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

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

\[x = \frac{3\pi - \pi}{6} + 2\pi n\]

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

\[Ответ:\ \ \frac{\pi}{3} + 2\pi n.\]

\[4)\sin{3x} + \cos{3x} = \sqrt{2}\ \ \ \ \ |\ :\sqrt{2}\]

\[\frac{\sqrt{2}}{2} \bullet \sin{3x} + \cos{3x} \bullet \frac{\sqrt{2}}{2} = 1\]

\[\cos\frac{\pi}{4} \bullet \sin{3x} + \cos{3x} \bullet \sin\frac{\pi}{4} = 1\]

\[\sin\left( \frac{\pi}{4} + 3x \right) = 1\]

\[3x + \frac{\pi}{4} = \arcsin 1 + 2\pi n\]

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

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

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

\[x = \frac{1}{3} \bullet \left( \frac{\pi}{4} + 2\pi n \right)\]

\[x = \frac{\pi}{12} + \frac{2\pi n}{3}.\]

\[Ответ:\ \ \frac{\pi}{12} + \frac{2\pi n}{3}.\]