Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 1207

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

\[+ 2 = 0\]

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

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

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

\[+ 2\left( \sin x + \cos x \right)^{2} -\]

\[- 4\sin x \bullet \cos x = 0\]

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

\[+ 2\left( \sin x + \cos x \right)^{2} -\]

\[- 2\sin{2x} = 0\]

\[Первое\ уравнение:\]

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

\[tg\ x + 1 = 0\]

\[tg\ x = - 1\]

\[x = - arctg\ 1 + \pi n = - \frac{\pi}{4} + \pi n.\ \]

\[Второе\ уравнение:\]

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

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

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

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

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

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

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

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

\[\left( \sin^{2}x + \cos^{2}x \right) + \sin{2x} + 2 -\]

\[- 3\left( \sin x + \cos x \right) = 0\]

\[\left( \sin x + \cos x \right)^{2} - 2\sin x \bullet\]

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

\[- 3\left( \sin x + \cos x \right) = 0\]

\[\left( \sin x + \cos x \right)^{2} -\]

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

\[+ \sin{2x} + 2 = 0\]

\[\left( \sin x + \cos x \right)^{2} -\]

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

\[Пусть\ y = \sin x + \cos x:\]

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

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

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

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

\[Первое\ уравнение:\]

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

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

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

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

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

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

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

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

\[Второе\ уравнение:\]

\[\sin x + \cos x = 2 - корней\ нет.\]

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

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

\[+ 4 = 0\]

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

\[+ 3 = 0\]

\[\sin{2x} + \left( \sin^{2}x + \cos^{2}x \right) +\]

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

\[\sin{2x} + \left( \sin x + \cos x \right)^{2} -\]

\[- 2\sin x \bullet \cos x +\]

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

\[\sin{2x} + \left( \sin x + \cos x \right)^{2} -\]

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

\[+ 3 = 0\]

\[\left( \sin x + \cos x \right)^{2} +\]

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

\[Пусть\ y = \sin x + \cos x:\]

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

\[D = 4^{2} - 4 \bullet 3 = 16 - 12 = 4\]

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

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

\[Первое\ уравнение:\]

\[\sin x + \cos x = - 3 - корней\]

\[\ нет.\]

\[Второе\ уравнение:\]

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

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

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

\[= - \frac{\sqrt{2}}{2}\]

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

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

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

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

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

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

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

\[= \pi + 2\pi n.\]

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

\[4)\sin{2x} +\]

\[+ 5\left( \cos x + \sin x + 1 \right) = 0\]

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

\[+ 4 = 0\]

\[\sin{2x} + \left( \cos^{2}x + \sin^{2}x \right) +\]

\[+ 5\left( \cos x + \sin x \right) + 4 = 0\]

\[\sin{2x} + \left( \sin x + \cos x \right)^{2} -\]

\[- 2\sin x \bullet \cos x +\]

\[+ 5\left( \cos x + \sin x \right) + 4 = 0\]

\[\sin{2x} + \left( \sin x + \cos x \right)^{2} -\]

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

\[+ 4 = 0\]

\[\left( \sin x + \cos x \right)^{2} +\]

\[+ 5\left( \cos x + \sin x \right) + 4 = 0\]

\[Пусть\ y = \sin x + \cos x:\]

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

\[D = 5^{2} - 4 \bullet 4 = 25 - 16 = 9\]

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

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

\[Первое\ уравнение:\]

\[\sin x + \cos x = - 4 - корней\ \]

\[нет.\]

\[Второе\ уравнение:\]

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

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

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

\[= - \frac{\sqrt{2}}{2}\]

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

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

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

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

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

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

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

\[= \pi + 2\pi n.\]

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