ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 671

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\[1)\sin\left( x + \frac{\pi}{6} \right) + \cos\left( x + \frac{\pi}{3} \right) =\]

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

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

\[y = \cos x:\]

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

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

\[y(2y - 1) = 0\]

\[y_{1} = 0\ \ и\ \ y_{2} = \frac{1}{2}.\]

\[1)\ \cos x = 0\]

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

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

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

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

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

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

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

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

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

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

\[1)\ \sin x = 0\]

\[x = \arcsin 0 + \pi n = \pi n.\]

\[2)\ \sqrt{2} - 2\cos x = 0\]

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

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

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

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

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