ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1477

\[\boxed{\mathbf{1477}\mathbf{.}}\]

\[1)\ y = \sin{2x} - \sqrt{3}\cos{2x} =\]

\[= 2 \bullet \left( \frac{1}{2}\sin{2x} - \frac{\sqrt{3}}{2}\cos{2x} \right) =\]

\[= 2 \bullet \left( \cos\frac{\pi}{3} \bullet \sin{2x} - \sin\frac{\pi}{3} \bullet \cos{2x} \right) =\]

\[= 2\sin\left( 2x - \frac{\pi}{3} \right);\]

\[- 1 \leq \sin\left( 2x - \frac{\pi}{3} \right) \leq 1\]

\[- 2 \leq 2\sin\left( 2x - \frac{\pi}{3} \right) \leq 2.\]

\[Ответ:\ \ \]

\[y_{\min} = - 2;\ \ y_{\max} = 2.\]

\[2)\ y = 2\cos{2x} + \sin^{2}x =\]

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

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

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

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

\[= 3\cos^{2}x - 1;\]

\[- 1 \leq \cos x \leq 1\]

\[0 \leq \cos^{2}x \leq 1\]

\[0 \leq 3\cos^{2}x \leq 3\]

\[- 1 \leq 3\cos^{2}x - 1 \leq 2.\]

\[Ответ:\ \ y_{\min} = - 1;\ \ y_{\max} = 2.\]