13 февраля 2017
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Задача 2
Решить неравенство
\(2 < {2^{{{\left( {\frac{{\sin x}}{{1 - \cos x}}} \right)}^2}}} < 8\)
Решение
ОДЗ: \(cos x \ne 1\).
Из условия имеем:
\(2 < {2^{{\rm{ct}}{{\rm{g}}^2}\frac{x}{2}}} < {2^3} \Leftrightarrow 1 < {\rm{ct}}{{\rm{g}}^2}\frac{x}{2} < 3 \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{ - \sqrt 3 < {\rm{ctg}}\frac{x}{2} < \sqrt {3,} }\\{\left[ {\begin{array}{*{20}{c}}{{\rm{ctg}}\frac{x}{2} > 1,}\\{{\rm{ctg}}\frac{x}{2} < - 1,}\end{array}} \right.}\end{array} \Leftrightarrow } \right.\)
\( \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{\frac{\pi }{6} + \pi n < \frac{x}{2} < \frac{{5\pi }}{6} + \pi n,}\\{\left[ {\begin{array}{*{20}{c}}{\pi n < \frac{x}{2} < \frac{\pi }{4} + \pi n,}\\{\frac{{3\pi }}{4} + \pi n < \frac{x}{2} < \pi + \pi n}\end{array}} \right.}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{\frac{\pi }{3} + 2\pi n < x < \frac{{5\pi }}{3} + 2\pi n,}\\{\left[ {\begin{array}{*{20}{c}}{2\pi n < x < \frac{\pi }{2} + 2\pi n,}\\{\frac{{3\pi }}{2} + 2\pi n < x < 2\pi + 2\pi n,}\end{array}} \right.}\end{array}\;\;\;n \in \mathbb{Z}.} \right.\)
Ответ: \(x \in \left( {\frac{\pi }{3} + 2\pi n;\;\frac{\pi }{2} + 2\pi n} \right) \cup \left( {\frac{3}{2}\pi + 2\pi n;\;\frac{5}{3}\pi + 2\pi n} \right),\;n \in \mathbb{Z}\)
\(2 < {2^{{{\left( {\frac{{\sin x}}{{1 - \cos x}}} \right)}^2}}} < 8\)
Решение
ОДЗ: \(cos x \ne 1\).
Из условия имеем:
\(2 < {2^{{\rm{ct}}{{\rm{g}}^2}\frac{x}{2}}} < {2^3} \Leftrightarrow 1 < {\rm{ct}}{{\rm{g}}^2}\frac{x}{2} < 3 \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{ - \sqrt 3 < {\rm{ctg}}\frac{x}{2} < \sqrt {3,} }\\{\left[ {\begin{array}{*{20}{c}}{{\rm{ctg}}\frac{x}{2} > 1,}\\{{\rm{ctg}}\frac{x}{2} < - 1,}\end{array}} \right.}\end{array} \Leftrightarrow } \right.\)
\( \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{\frac{\pi }{6} + \pi n < \frac{x}{2} < \frac{{5\pi }}{6} + \pi n,}\\{\left[ {\begin{array}{*{20}{c}}{\pi n < \frac{x}{2} < \frac{\pi }{4} + \pi n,}\\{\frac{{3\pi }}{4} + \pi n < \frac{x}{2} < \pi + \pi n}\end{array}} \right.}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{\frac{\pi }{3} + 2\pi n < x < \frac{{5\pi }}{3} + 2\pi n,}\\{\left[ {\begin{array}{*{20}{c}}{2\pi n < x < \frac{\pi }{2} + 2\pi n,}\\{\frac{{3\pi }}{2} + 2\pi n < x < 2\pi + 2\pi n,}\end{array}} \right.}\end{array}\;\;\;n \in \mathbb{Z}.} \right.\)
Ответ: \(x \in \left( {\frac{\pi }{3} + 2\pi n;\;\frac{\pi }{2} + 2\pi n} \right) \cup \left( {\frac{3}{2}\pi + 2\pi n;\;\frac{5}{3}\pi + 2\pi n} \right),\;n \in \mathbb{Z}\)