Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 661

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\[1)\ 6\sin^{2}x - \cos x + 6 = 0\]

\[6\left( 1 - \cos^{2}x \right) - \cos x + 6 = 0\]

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

\[6\cos^{2}x + \cos x - 12 = 0\]

\[y = \cos x:\]

\[6y^{2} + y - 12 = 0\]

\[D = 1 + 288 = 289\]

\[y_{1} = \frac{- 1 - 17}{2 \bullet 6} = - \frac{18}{12} = - \frac{3}{2};\]

\[y_{2} = \frac{- 1 + 17}{2 \bullet 6} = \frac{16}{12} = \frac{4}{3};\]

\[|y| > 1 - корней\ нет.\]

\[Ответ:\ \ корней\ нет\]

\[2)\ 8\cos^{2}x - 12\sin x + 7 = 0\]

\[8\left( 1 - \sin^{2}x \right) - 12\sin x + 7 = 0\]

\[8 - 8\sin^{2}x - 12\sin x + 7 = 0\]

\[8\sin^{2}x + 12\sin x - 15 = 0\]

\[Пусть\ y = \sin x,\ тогда:\]

\[8y^{2} + 12y - 15 = 0\]

\[D = 144 + 480 = 624 = 16 \bullet 39\]

\[y = \frac{- 12 \pm 4\sqrt{39}}{2 \bullet 8} = \frac{- 3 \pm \sqrt{39}}{4}.\]

\[1)\ \sin x = \frac{- 3 - \sqrt{39}}{4}\]

\[корней\ нет.\]

\[2)\ \sin x = \frac{- 3 + \sqrt{39}}{4}\]

\[x = ( - 1)^{n} \bullet \arcsin\frac{\sqrt{39} - 3}{4} + \pi n.\]

\[Ответ:\ \ \]

\[( - 1)^{n} \bullet \arcsin\frac{\sqrt{39} - 3}{4} + \pi n.\]