$$\begin{cases} \cos x-\sqrt{3}\sin x = \sqrt{3}\\ \sqrt{1+\sin x} = -\cos x\\ \sqrt{1-\sin x} = -\cos x\\ \end{cases}$$ задан 18 Апр '12 22:17 isuk9696 |
$% \begin {cases} x \in \mathbb{R} \wedge \cos x - \sqrt{3} \sin x = \sqrt{3} \wedge \sqrt{1 + \sin x} = - \cos x \wedge \sqrt{1 - \sin x} = - \cos x \\ 1 + \sin x \geq 0 \wedge 1 - \sin x \geq 0 \end {cases}$% $% \Leftrightarrow \begin {cases} x \in \mathbb{R} \wedge \sin x \geq - 1 \wedge \sin x \leq 1 \\ \cos x = \sqrt{3} \cdot (1 + \sin x) \wedge \sqrt{1 + \sin x} = - \cos x \wedge - \cos x = \sqrt{1 - \sin x} \end {cases} $% $% \Rightarrow x \in \mathbb{R} \wedge - 1 \leq \sin x \leq 1 \wedge \cos x = \sqrt{3}(1 + \sin x) \wedge \sqrt{1 + \sin x} = \sqrt{1 - \sin x} $% $% \Rightarrow x \in \mathbb{R} \wedge \sin x \in [-1, 1] \wedge \cos x = \sqrt{3}(1 + \sin x) \wedge (\sqrt{1 + \sin x})^2 = (\sqrt{1 - \sin x})^2 $% $% \Leftrightarrow x \in \mathbb{R} \wedge \sin x \in [-1, 1] \wedge \cos x = \sqrt{3}(1 + \sin x) \wedge 2 \sin x = 0 $% $% \Leftrightarrow x \in \mathbb{R} \wedge \sin x \in [-1, 1] \wedge \cos x = \sqrt{3}(1 + \sin x) \wedge \sin x = 0 $% $% \Rightarrow x \in \mathbb{R} \wedge \cos x = \sqrt{3}(1 + 0) \wedge \mathrm{True} $% $% \Rightarrow x \in \mathbb{R} \wedge \cos x = \sqrt{3} \wedge (x \in \mathbb{R} \rightarrow -1 \leq \cos x \leq 1) $% $% \Leftrightarrow x \in \mathbb{R} \wedge \cos x = \sqrt{3} \wedge -1 \leq \cos x \leq 1 $% $% \Rightarrow x \in \mathbb{R} \wedge \cos x \in \{\sqrt{3}\} \wedge \cos x \in [-1, 1] $% $% \Rightarrow x \in \mathbb{R} \wedge \cos x \in \{\sqrt{3}\} \cap [-1, 1] $% $% \Rightarrow x \in \mathbb{R} \wedge \cos x \in \varnothing $% $% \Leftrightarrow False $% отвечен 20 Апр '12 10:35 Галактион |