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МатематикаРешебник по алгебре и начала математического анализа 10 класс Колягин Задание 1193
Задание 1193
\[\boxed{\mathbf{1193}\mathbf{.}}\]
\[1)\ 2\cos^{2}x - \sin x + 1 = 0\]
\[2\left( 1 - \sin^{2}x \right) - \sin x + 1 = 0\]
\[2 - 2\sin^{2}x - \sin x + 1 = 0\]
\[2\sin^{2}x + \sin x - 3 = 0\]
\[Пусть\ y = \sin x:\]
\[2y^{2} + y - 3 = 0\]
\[D = 1^{2} + 4 \bullet 2 \bullet 3 = 1 + 24 = 25\]
\[y_{1} = \frac{- 1 - 5}{2 \bullet 2} = - \frac{3}{2}\text{\ \ }и\ \ \]
\[y_{2} = \frac{- 1 + 5}{2 \bullet 2} = 1\]
\[Первое\ уравнение:\]
\[\sin x = - \frac{3}{2} - корней\ нет.\]
\[Второе\ уравнение:\]
\[\sin x = 1\]
\[x = \arcsin 1 + 2\pi n = \frac{\pi}{2} + 2\pi n.\]
\[Ответ:\ \ \frac{\pi}{2} + 2\pi n.\]
\[2)\ 3\cos^{2}x - \sin x - 1 = 0\]
\[3\left( 1 - \sin^{2}x \right) - \sin x - 1 = 0\]
\[3 - 3\sin^{2}x - \sin x - 1 = 0\]
\[3\sin^{2}x + \sin x - 2 = 0\]
\[Пусть\ y = \sin x:\]
\[3y^{2} + y - 2 = 0\]
\[D = 1^{2} + 4 \bullet 3 \bullet 2 = 1 + 24 = 25\]
\[y_{1} = \frac{- 1 - 5}{2 \bullet 3} = - 1\ \ и\]
\[\text{\ \ }y_{2} = \frac{- 1 + 5}{2 \bullet 3} = \frac{2}{3}.\]
\[Первое\ уравнение:\]
\[\sin x = - 1\]
\[x = - \arcsin 1 + 2\pi n =\]
\[= - \frac{\pi}{2} + 2\pi n.\]
\[Второе\ уравнение:\]
\[\sin x = \frac{2}{3}\]
\[x = ( - 1)^{n} \bullet \arcsin\frac{2}{3} + \pi n.\]
\[Ответ:\ - \frac{\pi}{2} + 2\pi n;\ \ \]
\[( - 1)^{n} \bullet \arcsin\frac{2}{3} + \pi n.\]
\[3)\ 4\sin^{2}x - \cos x - 1 = 0\]
\[4\left( 1 - \cos^{2}x \right) - \cos x - 1 = 0\]
\[4 - 4\cos^{2}x - \cos x - 1 = 0\]
\[4\cos^{2}x + \cos x - 3 = 0\]
\[Пусть\ y = \cos x:\]
\[4y^{2} + y - 3 = 0\]
\[D = 1^{2} + 4 \bullet 4 \bullet 3 = 1 + 48 = 49\]
\[y_{1} = \frac{- 1 - 7}{2 \bullet 4} = - 1\ \ и\ \]
\[\ y_{2} = \frac{- 1 + 7}{2 \bullet 4} = \frac{3}{4}.\]
\[Первое\ уравнение:\]
\[\cos x = - 1\]
\[x = \pi - \arccos 1 + 2\pi n =\]
\[= \pi + 2\pi n.\]
\[Второе\ уравнение:\]
\[\cos x = \frac{3}{4}\]
\[x = \pm \arccos\frac{3}{4} + 2\pi n.\]
\[Ответ:\ \ \pi + 2\pi n;\]
\[\ \ \pm \arccos\frac{3}{4} + 2\pi n.\]
\[4)\ 2\sin^{2}x + 3\cos x = 0\]
\[2\left( 1 - \cos^{2}x \right) + 3\cos x = 0\]
\[2 - 2\cos^{2}x + 3\cos x = 0\]
\[2\cos^{2}x - 3\cos x - 2 = 0\]
\[Пусть\ y = \cos x:\]
\[2y^{2} - 3y - 2 = 0\]
\[D = 3^{2} + 4 \bullet 2 \bullet 2 = 9 + 16 = 25\]
\[y_{1} = \frac{3 - 5}{2 \bullet 2} = - \frac{1}{2}\text{\ \ }и\ \]
\[\ y_{2} = \frac{3 + 5}{2 \bullet 2} = 2.\]
\[Первое\ уравнение:\]
\[\cos x = - \frac{1}{2}\]
\[x = \pm \left( \pi - \arccos\frac{1}{2} \right) + 2\pi n =\]
\[= \pm \left( \pi - \frac{\pi}{3} \right) + 2\pi n =\]
\[= \pm \frac{2\pi}{3} + 2\pi n.\]
\[Второе\ уравнение:\]
\[\cos x = 2 - корней\ нет.\]
\[Ответ:\ \pm \frac{2\pi}{3} + 2\pi n.\]
