ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1382

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

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

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

\[D = 25 - 16 = 9\]

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

\[y_{2} = \frac{5 + 3}{2} = 4.\]

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

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

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

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

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

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

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

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

\[\sin x + \cos x = 4\]

\[корней\ нет.\]

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

\[2)\ 2 + 2\cos x = 3\sin x \bullet \cos x + 2\sin x\]

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

\[3y^{2} + 4y + 1 = 0\]

\[D = 16 - 12 = 4\]

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

\[y_{2} = \frac{- 4 + 2}{2 \bullet 3} = - \frac{1}{3}.\]

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

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

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

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

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

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

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

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

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

\[= - \pi + 2\pi n = \pi + 2\pi n;\]

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

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

\[2)\ \cos x - \sin x = - \frac{1}{3}\ \ \ \ \ |\ :\sqrt{2}\]

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

\[\sin\frac{\pi}{4} \bullet \cos x - \cos\frac{\pi}{4} \bullet \sin x = - \frac{1}{3\sqrt{2}}\]

\[\sin\left( \frac{\pi}{4} - x \right) = - \frac{1}{3\sqrt{2}}\]

\[\sin\left( x - \frac{\pi}{4} \right) = \frac{1}{3\sqrt{2}}\]

\[x - \frac{\pi}{4} = ( - 1)^{n} \bullet \arcsin\frac{1}{3\sqrt{2}} + \pi n\]

\[x = \frac{\pi}{4} + ( - 1)^{n} \bullet \arcsin\frac{1}{3\sqrt{2}} + \pi n.\]

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

\[\frac{\pi}{4} + ( - 1)^{n} \bullet \arcsin\frac{1}{3\sqrt{2}} + \pi n.\]