Страница 431 номер 1624, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

\(y = \dfrac{2\cos^{4}x + \sin^{2}x}{2\sin^{4}x + 3\cos^{2}x} = \dfrac{2\cos^{4}x + 1 - \cos^{2}x}{2\sin^{4}x + 3 - 3\sin^{2}x} =\)
\(= \dfrac{\cos^{2}x\left( 2\cos^{2}x - 1 \right) + 1}{\sin^{2}x\left( 2\sin^{2}x - 3 \right) + 3} =\)
\(= \dfrac{\cos^{2}x\left( \cos^{2}x - \sin^{2}x \right) + 1}{\sin^{2}x\left(-\sin^{2}x - 3\cos^{2}x \right) + 3} =\)
\(= \dfrac{\cos^{2}x\cos{2x} + 1}{\sin^{2}x\left(-1 - 2\cos^{2}x \right) + 3} = \dfrac{\cos^{2}x\cos{2x} + 1}{-\sin^{2}x - 2\sin^{2}x\cos^{2}x + 3} =\)
\(= \dfrac{\cos^{2}x\cos{2x} + 1}{2 + \cos^{2}x - 2\sin^{2}x\cos^{2}x} = \dfrac{\cos^{2}x\cos{2x} + 1}{2 + \cos^{2}x\left( 1 - 2\sin^{2}x \right)} =\)
\(= \dfrac{\cos^{2}x\cos{2x} + 1}{2 + \cos^{2}x\left( \cos^{2}x - \sin^{2}x \right)} =\)
\(= \dfrac{\cos^{2}x\cos{2x} + 1}{\cos^{2}x\cos{2x} + 2},\)
\(y'(x) = \dfrac{\left(-2\cos x\cos{2x}\sin x - 2\cos^{2}x\sin{2x} \right)\left( \cos^{2}x\cos{2x} + 2 \right)}{\left( \cos^{2}x\cos{2x} + 2 \right)^{2}} -\)
\(- \dfrac{\left( \cos^{2}x\cos{2x} + 1 \right)\left(-2\cos x\cos{2x}\sin x - 2\cos^{2}x\sin{2x} \right)}{\left( \cos^{2}x\cos{2x} + 2 \right)^{2}} =\)
\(= \dfrac{\begin{gathered}\left(-\sin{2x}\cos{2x} - 2\cos^{2}x\sin{2x} \right) \cdot {} \\ {} \cdot \left( \cos^{2}x\cos{2x} + 2 - \cos^{2}x\cos{2x} - 1 \right)\end{gathered}}{\left( \cos^{2}x\cos{2x} + 2 \right)^{2}} =\)
\(= \dfrac{-\sin{2x}\left( \cos{2x} + 2\cos^{2}x \right)}{\left( \cos^{2}x\cos{2x} + 2 \right)^{2}},\)
\(\left( \cos^{2}x\cos{2x} + 2 \right)^{2} > 0,\)
\(- \sin{2x}\left( 4\cos^{2}x - 1 \right) = 0,\)
\(\sin{2x} = 0\) или \(4\cos^{2}x - 1 = 0\)
\(x_{1} = \frac{\pi n}{2}\), \(n \in Z\) – точка максимума;
\(x_{2} = \pm \frac{\pi}{3} + \pi n\), \(n \in Z\) – точка минимума;
\(y(\pi) = \dfrac{2}{3}, y\left( \dfrac{\pi}{2} \right) = \dfrac{1}{2};\)
\(y_{\max} = y(\pi) = \dfrac{2}{3};\)
\(y_{\min} = y\left( \dfrac{\pi}{3} \right) = \dfrac{7}{15}.\)