Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 647

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

\[(1 - a) \bullet tg^{2}\ x - tg\ x - (2 + a) = 0\]

\[y = tg\ x:\]

\[(1 - a)y^{2} - y - (2 + a) = 0\]

\[D = 1^{2} + 4(1 - a)(2 + a) =\]

\[= 1 + 4\left( 2 + a - 2a - a^{2} \right) =\]

\[= 1 + 8 - 4a - 4a^{2}\]

\[При\ D < 0:\]

\[9 - 4a - 4a^{2} < 0\]

\[4a^{2} + 4a - 9 > 0\]

\[D_{2} = 16 + 144 = 160 = 16 \bullet 10\]

\[a = \frac{- 4 \pm 4\sqrt{10}}{4 \bullet 2} = \frac{- 1 \pm \sqrt{10}}{2}.\]

\[\left( a - \frac{- 1 - \sqrt{10}}{2} \right)\left( a + \frac{- 1 + \sqrt{10}}{2} \right) > 0\]

\[a_{1} < \frac{- 1 - \sqrt{10}}{2};\]

\[a_{2} > \frac{- 1 + \sqrt{10}}{2}.\]

\[Ответ:\ \ a < - \frac{1 + \sqrt{10}}{2};\]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }a > \frac{\sqrt{10} - 1}{2}.\]