ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1245

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

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

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

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

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

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

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

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

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

\[Пусть\ y = \cos x:\]

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

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

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

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

\[Первое\ уравнение:\]

\[\cos x = 0\]

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

\[Второе\ уравнение:\]

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

\[x = \pm \arccos\frac{1}{2} + 2\pi n = \pm \frac{\pi}{3} +\]

\[+ 2\pi n.\]

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

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

\[= \sin{2x}\]

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

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

\[= \sin{2x}\]

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

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

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

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

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

\[Первое\ уравнение:\]

\[\sin x = 0\]

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

\[Второе\ уравнение:\]

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

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

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

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

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

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