Nierówność trygonometryczna

[Iluminati91]

\( \frac{1-2 \sin x}{ \cos 2x} \ge 0 \) dla \( 0<x<\Pi\)

[eresh]

\( \frac{1-2 \sin x}{ \cos 2x} \ge 0 \) dla \( 0<x<\Pi\)

\(\frac{1-2\sin x}{\cos 2x}\ge 0\\
\cos 2x\neq 0\\
2x\neq \frac{\pi}{2}+k\pi\\
x\neq\frac{\pi}{4}+\frac{k\pi}{2}\\
x\neq \frac{\pi}{4},\;\;x\neq \frac{3\pi}{4}\)

\((1-2\sin x)\cos 2x\geq 0\\
(1-2\sin x)(1-2\sin ^2x)\geq 0\\
\sin x=t\\
(1-2t)(1-2t^2)\geq 0\\
t\in (-\frac{\sqrt{2}}{2},\frac{1}{2}]\cup (\frac{\sqrt{2}}{2},\infty)\\
-\frac{\sqrt{2}}{2}<\sin x\leq \frac{1}{2}\;\;\; \vee \;\;\;\sin x>\frac{\sqrt{2}}{2}\\
x\in (0,\frac{\pi}{6}]\cup(\frac{\pi}{4},\frac{3\pi}{4})\cup [\frac{5\pi}{6},\pi)\)