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МатематикаГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1195
Задание 1195
\[\boxed{\mathbf{1195}\mathbf{.}}\]
\[1)\ 1 + 7\cos^{2}x = 3\sin{2x}\]
\[\cos^{2}x + \sin^{2}x + 7\cos^{2}x =\]
\[= 3 \bullet 2\sin x \bullet \cos x\]
\[\sin^{2}x + 8\cos^{2}x -\]
\[- 6\sin x \bullet \cos x = 0\ \ \ \ \ |\ :\cos^{2}x\]
\[tg^{2}x + 8 - 6\ tg\ x = 0\]
\[Пусть\ y = tg\ x:\]
\[y^{2} - 6y + 8 = 0\]
\[D = 6^{2} - 4 \bullet 8 = 36 - 32 = 4\]
\[y_{1} = \frac{6 - 2}{2} = 2\ \ и\]
\[\text{\ \ }y_{2} = \frac{6 + 2}{2} = 4.\]
\[Первое\ уравнение:\]
\[tg\ x = 2\]
\[x = arctg\ 2 + \pi n.\]
\[Второе\ уравнение:\]
\[tg\ x = 4\]
\[x = arctg\ 4 + \pi n.\]
\[Ответ:\ \ arctg\ 2 + \pi n;\ \ \]
\[arctg\ 4 + \pi n.\]
\[2)\ 3 + \sin{2x} = 4\sin^{2}x\]
\[3\left( \cos^{2}x + \sin^{2}x \right) +\]
\[+ 2\sin x \bullet \cos x = 4\sin^{2}x\]
\[3\cos^{2}x + 3\sin^{2}x +\]
\[+ 2\sin x \bullet \cos x - 4\sin^{2}x\]
\[3\cos^{2}x - \sin^{2}x +\]
\[+ 2\sin x \bullet \cos x = 0\ \ \ \ \ |\ :\cos^{2}x\]
\[3 - tg^{2}\ x + 2\ tg\ x = 0\]
\[Пусть\ y = tg\ x:\]
\[3 - y^{2} + 2y = 0\]
\[y^{2} - 2y - 3 = 0\]
\[D = 2^{2} + 4 \bullet 3 = 4 + 12 = 16\]
\[y_{1} = \frac{2 - 4}{2} = - 1\ \ и\ \ \]
\[y_{2} = \frac{2 + 4}{2} = 3.\]
\[Первое\ уравнение:\]
\[tg\ x = - 1\]
\[x = - arctg\ 1 + \pi n =\]
\[= - \frac{\pi}{4} + \pi n.\]
\[Второе\ уравнение:\]
\[tg\ x = 3\]
\[x = arctg\ 3 + \pi n.\]
\[Ответ:\ - \frac{\pi}{4} + \pi n;\ \ arctg\ 3 + \pi n.\]
\[3)\cos{2x} + \cos^{2}x +\]
\[+ \sin x \bullet \cos x = 0\]
\[\cos^{2}x - \sin^{2}x + \cos^{2}x +\]
\[+ \sin x \bullet \cos x = 0\]
\[2\cos^{2}x - \sin^{2}x + \sin x \bullet \cos x =\]
\[= 0\ \ \ \ \ |\ :\cos^{2}x\]
\[2 - tg^{2}\ x + tg\ x = 0\]
\[Пусть\ y = tg\ x:\]
\[2 - y^{2} + y = 0\]
\[y^{2} - y - 2 = 0\]
\[D = 1^{2} + 4 \bullet 2 = 1 + 8 = 9\]
\[y_{1} = \frac{1 - 3}{2} = - 1\ \ и\ \]
\[\ y_{2} = \frac{1 + 3}{2} = 2.\]
\[Первое\ уравнение:\]
\[tg\ x = - 1\]
\[x = - arctg\ 1 + \pi n = - \frac{\pi}{4} + \pi n.\]
\[Второе\ уравнение:\]
\[tg\ x = 2\]
\[x = arctg\ 2 + \pi n.\]
\[Ответ:\ - \frac{\pi}{4} + \pi n;\ \ arctg\ 2 + \pi n.\]
\[4)\ 3\cos{2x} + \sin^{2}x +\]
\[+ 5\sin x \bullet \cos x = 0\]
\[3\left( \cos^{2}x - \sin^{2}x \right) + \sin^{2}x +\]
\[+ 5\sin x \bullet \cos x = 0\]
\[3\cos^{2}x - 3\sin^{2}x + \sin^{2}x +\]
\[+ 5\sin x \bullet \cos x = 0\]
\[3\cos^{2}x - 2\sin^{2}x +\]
\[+ 5\sin x \bullet \cos x = 0\ \ \ \ \ |\ :\cos^{2}x\]
\[3 - 2\ tg^{2}x + 5\ tg\ x = 0\]
\[Пусть\ y = tg\ x:\]
\[3 - 2y^{2} + 5y = 0\]
\[2y^{2} - 5y - 3 = 0\]
\[D = 5^{2} + 4 \bullet 2 \bullet 3 =\]
\[= 25 + 24 = 49\]
\[y_{1} = \frac{5 - 7}{2 \bullet 2} = - \frac{1}{2}\text{\ \ }и\ \ \]
\[y_{2} = \frac{5 + 7}{2 \bullet 2} = 3.\]
\[Первое\ уравнение:\]
\[tg\ x = - \frac{1}{2}\]
\[x = - arctg\frac{1}{2} + \pi n.\]
\[Второе\ уравнение:\]
\[tg\ x = 3\]
\[x = arctg\ 3 + \pi n.\]
\[Ответ:\ - arctg\frac{1}{2} + \pi n;\ \]
\[\ arctg\ 3 + \pi n.\]
