Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 631

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\[1)\ \sin x + \cos x = 0\ \ \ \ \ |\ :\cos x\]

\[tg\ x + 1 = 0\]

\[tg\ x = - 1\]

\[x = - arctg\ 1 + \pi n\]

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

\[2)\ 2\sin x + 2\cos x - 3 = 0\]

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

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

\[\cos\frac{\pi}{4} \bullet \sin x + \cos x \bullet \sin\frac{\pi}{4} = \frac{3}{2\sqrt{2}}\]

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

\[нет\ корней.\]

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

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

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

\[y = \sin x + \cos x:\]

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

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

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

\[y_{2} = \frac{3 + 1}{2} = 2.\]

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

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

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

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

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

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

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

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

\[\sin x + \cos x = 2\]

\[корней\ нет.\]

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

\[3)\sin{2x} + 4\left( \sin x + \cos x \right) + 4 = 0\]

\[\sin{2x} + 1 + 4\left( \sin x + \cos x \right) + 3 = 0\]

\[y = \sin x + \cos x:\]

\[y^{2} + 4y + 3 = 0\]

\[D = 16 - 12 = 4\]

\[y_{1} = \frac{- 4 - 2}{2} = - 3;\]

\[y_{2} = \frac{- 4 + 2}{2} = - 1.\]

\[1)\ \sin x + \cos x = - 3\]

\[корней\ нет.\]

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

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

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

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

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

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

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

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

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

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

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

\[4)\sin{2x} + 5\left( \cos x + \sin x + 1 \right) = 0\]

\[\sin{2x} + 1 + 5\left( \cos x + \sin x \right) + 4 = 0\]

\[y = \sin x + \cos x:\]

\[y^{2} + 5y + 4 = 0\]

\[D = 25 - 16 = 9\]

\[y_{1} = \frac{- 5 - 3}{2} = - 4;\]

\[y_{2} = \frac{- 5 + 3}{2} = - 1.\]

\[1)\ \sin x + \cos x = - 4\]

\[корней\ нет.\]

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

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

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

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

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

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

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

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

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

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

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