2010014901

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2010014901
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Řešením nerovnice \( \sin x \geq \frac{\sqrt{2}}2 \) pro \( x\in\mathbb{R} \) je množina:
\( \bigcup\limits_{k\in\mathbb{Z}}\left\langle\frac{\pi}4+2k\pi;\ \frac{3\pi}4+2k\pi\right\rangle \)
\( \bigcup\limits_{k\in\mathbb{Z}}\left\langle \frac{\pi}4+k\pi;\ \frac{3\pi}4+k\pi\right\rangle \)
\( \bigcup\limits_{k\in\mathbb{Z}}\left\langle -\frac{\pi}4+2k\pi;\ \frac{\pi}4+2k\pi\right\rangle \)
\( \bigcup\limits_{k\in\mathbb{Z}}\left\langle -\frac{\pi}4+k\pi;\ \frac{\pi}4+k\pi\right\rangle \)