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Question Number 103859 by 175mohamed last updated on 17/Jul/20

$$$$$$solve:$$$$y^2+x^2=-\sin \left(\theta \right).,\frac{\pi }{2}⩾\theta ⩾-\frac{\pi }{2}$$

Answered by mathmax by abdo last updated on 18/Jul/20

$$\mathrm{if}\mathrm{x}\mathrm{and}\mathrm{y}\mathrm{from}\mathrm{C}\mathrm{let}\mathrm{x}=\mathrm{rcos}\alpha \mathrm{and}\mathrm{y}=\mathrm{r}\sin \alpha$$$$\mathrm{e}\Rightarrow {\mathrm{r}}^2=-\sin \theta \mathrm{but}\mathrm{r}>0\left(\mathrm{radius}\right)\Rightarrow \sin \theta <0\Rightarrow -\frac{\pi }{2}<\theta <0\Rightarrow$$$$\mathrm{r}=\sqrt{-\sin \theta }\Rightarrow \mathrm{x}=\sqrt{-\sin \theta }\cos \left(\alpha \right)\mathrm{and}\mathrm{y}=\sqrt{-\sin \theta }\sin \left(\alpha \right)\mathrm{and}\in ]0,2\pi [$$$$\mathrm{if}\sin \theta >0\to \mathrm{no}\mathrm{solution}...$$

Commented by mathmax by abdo last updated on 18/Jul/20

$$\mathrm{sorry}\mathrm{x}\mathrm{and}\mathrm{y}\mathrm{from}\mathrm{R}..$$

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