ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1243

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

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

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

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

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

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

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

\[- 3\sin^{2}x = 0\ \ \ \ \ |\ :\cos^{2}x\]

\[2\ tg\ x + 1 - 3\ tg^{2}\ x = 0\]

\[Пусть\ y = tg\ x:\]

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

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

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

\[y_{1} = \frac{2 - 4}{2 \bullet 3} = - \frac{2}{6} = - \frac{1}{3}\text{\ \ }и\ \]

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

\[Первое\ уравнение:\]

\[tg\ x = - \frac{1}{3}\]

\[x = - arctg\frac{1}{3} + \pi n.\]

\[Второе\ уравнение:\]

\[tg\ x = 1\]

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

\[Ответ:\ - arctg\frac{1}{3} + \pi n;\ \ \frac{\pi}{4} + \pi n.\]

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

\[\cos^{2}x - \sin^{2}x + 3 \bullet 2\sin x \bullet\]

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

\[\cos^{2}x - 3\cos^{2}x - \sin^{2}x -\]

\[- 3\sin^{2}x + 6\sin x \bullet \cos x = 0\]

\[- 2\cos^{2}x - 4\sin^{2}x + 6\sin x \bullet\]

\[\bullet \cos x = 0\ \ \ \ \ |\ :\cos^{2}x\]

\[- 2 - 4\ tg^{2}\ x + 6\ tg\ x = 0\]

\[Пусть\ y = tg\ x:\]

\[- 2 - 4y^{2} + 6y = 0\]

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

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

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

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

\[Первое\ уравнение:\]

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

\[x = arctg\frac{1}{2} + \pi n.\]

\[Второе\ уравнение:\]

\[tg\ x = 1\]

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

\[Ответ:\ \ arctg\frac{1}{2} + \pi n;\ \ \frac{\pi}{4} + \pi n.\]