ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 646

\[\boxed{\mathbf{646}\mathbf{.}}\]

\[y = \cos x:\]

\[4y^{2} - 2(a - 3)y - 3a = 0\]

\[D = 4(a - 3)^{2} + 4 \bullet 4 \bullet 3a =\]

\[= 4a^{2} - 24a + 9 + 48a =\]

\[= 4a^{2} + 24a + 9 = 4(a + 3)^{2}\]

\[y = \frac{2(a - 3) \pm 2(a + 3)}{2 \bullet 4} =\]

\[= \frac{a - 3 \pm (a + 3)}{4};\]

\[y_{1} = \frac{a - 3 - a - 3}{4} = - \frac{6}{4} = - \frac{3}{2};\]

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

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

\[нет\ корней.\]

\[2)\ \cos x = \frac{a}{2}\]

\[\left| \frac{a}{2} \right| \leq 1\]

\[|a| \leq 2\]

\[x = \pm \arccos\frac{a}{2} + 2\pi n.\]

\[Ответ:\ \ |a| \leq 2;\ \ \]

\[x = \pm \arccos\frac{a}{2} + 2\pi n.\]