解決r=sqrt{1+cos2θ} |Microsoft數學求解器

$$r = \sqrt { 1 + \cos 2 \theta }$$

解 r (復數求解)

$$r=\sqrt{\cos(2\theta )+1}$$

解 θ (復數求解)

$$\left\{\begin{matrix}\theta =\pi n_{1}+\frac{\pi }{2}\text{, }n_{1}\in \mathrm{Z}\text{, }&r=0\\\theta =-\frac{i\ln(-\sqrt{r^{2}\left(r^{2}-2\right)}+r^{2}-1)}{2}+\pi n_{2}\text{, }n_{2}\in \mathrm{Z}\text{; }\theta =-\frac{i\ln(\sqrt{r^{2}\left(r^{2}-2\right)}+r^{2}-1)}{2}+\pi n_{3}\text{, }n_{3}\in \mathrm{Z}\text{, }&arg(r)<\pi \text{ and }r\neq 0\end{matrix}\right.$$

解 r

$$r=\sqrt{2}|\cos(\theta )|$$

解 θ

$$\theta =\pi n_{1}+\frac{\pi }{2}$$

$$n_{1}\in \mathrm{Z}$$

$$r=0$$