Тригонометријске неједначине. Једноставни примери.
Текст задатака објашњених у видео лекцији.
Пр.1 $2\sin x - \sqrt 2 > 0$
Пр.2 $2\cos x + 1 > 0$
Пр.3 $tgx + 1 < 0$
Пр.4 $ - 2\sin x - \sqrt 3 < 0$
Пр.5 $\sin \left( {\frac{{3\pi }}{2} - x} \right) > \frac{{\sqrt 3 }}{2}$
Пр.1
\[\begin{gathered}
2\sin x - \sqrt 2 > 0 \hfill \\
2\sin x > \sqrt 2 \hfill \\
\sin x > \frac{{\sqrt 2 }}{2} \hfill \\
x \in \left( {\frac{\pi }{4} + 2k\pi ,\frac{{3\pi }}{4} + 2k\pi } \right),k \in \mathbb{Z} \hfill \\
\end{gathered} \]
Пр.2
\[\begin{gathered}
2\cos x + 1 > 0 \hfill \\
2\cos x > - 1 \hfill \\
\cos x > - \frac{1}{2} \hfill \\
x \in \left( {0 + 2k\pi ,\frac{{2\pi }}{3} + 2k\pi } \right) \cup \left( {\frac{{4\pi }}{3} + 2k\pi ,2\pi + 2k\pi } \right),k \in \mathbb{Z} \hfill \\
\end{gathered} \]
Пр.3
\[\begin{gathered}
tgx + 1 < 0 \hfill \\
tgx < - 1 \hfill \\
x \in \left( {\frac{\pi }{2} + k\pi ,\frac{{3\pi }}{4} + k\pi } \right),k \in \mathbb{Z} \hfill \\
\hfill \\
\end{gathered} \]
Пр.4
\[\begin{gathered}
- 2\sin x - \sqrt 3 < 0 \hfill \\
- 2\sin x < \sqrt 3 \hfill \\
\sin x > - \frac{{\sqrt 3 }}{2} \hfill \\
x \in \left( {0 + 2k\pi ,\frac{{4\pi }}{3} + 2k\pi } \right) \cup \left( {\frac{{5\pi }}{3} + 2k\pi ,2\pi + 2k\pi } \right),k \in \mathbb{Z} \hfill \\
\end{gathered} \]
Пр.5
$\left( {\frac{{3\pi }}{2} - x} \right) > \frac{{\sqrt 3 }}{2}$
$\frac{\pi }{3} + 2k\pi < \frac{{3\pi }}{2} - x < \frac{{2\pi }}{3} + 2k\pi $
$\frac{{3\pi }}{2} - x > \frac{\pi }{3} + 2k\pi $ |
$ \cap $ |
$\frac{{3\pi }}{2} - x < \frac{{2\pi }}{3} + 2k\pi $ |
$ - x > \frac{\pi }{3} - \frac{{3\pi }}{2} + 2k\pi $ |
|
$ - x < \frac{{2\pi }}{3} - \frac{{3\pi }}{2} + 2k\pi $ |
$ - x > - \frac{{7\pi }}{6} + 2k\pi \left| { \cdot \left( { - 1} \right)} \right.$ |
|
$ - x < - \frac{{5\pi }}{6} + 2k\pi \left| { \cdot \left( { - 1} \right)} \right.$ |
$x < \frac{{7\pi }}{6} - 2k\pi ,k \in \mathbb{Z}$ |
|
$x > \frac{{5\pi }}{6} - 2k\pi ,k \in \mathbb{Z}$ |
\[x \in \left( {\frac{{5\pi }}{6} - 2k\pi ;\frac{{7\pi }}{6} - 2k\pi } \right),k \in \mathbb{Z}\]