Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 1268

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\[\cos x - \sin x - \cos{2x} > 0;\]

\[\text{\ \ \ \ }(0;2\pi)\text{.\ \ }\]

\[f(x) = \cos x - \sin x -\]

\[- \cos{2x} > 0\]

\[\cos x - \sin x -\]

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

\[\left( \cos x - \sin x \right) -\]

\[- \left( \cos x - \sin x \right)\left( \cos x + \sin x \right) =\]

\[= 0\]

\[(\cos x - \sin x\left( 1 - \cos x - \sin x \right) = 0\]

\[1)\ \cos x - \sin x = 0\]

\[1 - tg\ x = 0\]

\[tg\ x = 1\]

\[x = \frac{\pi}{4} + \pi n.\]

\[2)\ 1 - \sqrt{2}\sin\left( x + \frac{\pi}{4} \right) = 0\]

\[\sin\left( x + \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2}\]

\[x + \frac{\pi}{4} = ( - 1)^{n} \cdot \frac{\pi}{4} + \pi n\]

\[x = - \frac{\pi}{4} + ( - 1)^{n} \cdot \frac{\pi}{4} + \pi n.\]

\[\frac{\pi}{4} < x < \frac{\pi}{2};\ \ \ \frac{5\pi}{4} < x < 2\pi.\]