求解 x^2+y^2=100cos(θ)

$${ x }^{ 2 } + { y }^{ 2 } =100 \cos ( \theta )$$

求解 x 的值 (复数求解)

$$x=-\sqrt{100\cos(\theta )-y^{2}}$$

$$x=\sqrt{100\cos(\theta )-y^{2}}$$

求解 y 的值 (复数求解)

$$y=-\sqrt{100\cos(\theta )-x^{2}}$$

$$y=\sqrt{100\cos(\theta )-x^{2}}$$

求解 x 的值

$$x=\sqrt{100\cos(\theta )-y^{2}}$$

$$x=-\sqrt{100\cos(\theta )-y^{2}}\text{, }\exists n_{1}\in \mathrm{Z}\text{ : }\left(\theta \geq \frac{\pi \left(4n_{1}+3\right)}{2}\text{ and }\theta \leq \frac{\pi \left(4n_{1}+5\right)}{2}\right)\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }\left(\theta \geq \frac{\pi \left(4n_{1}-1\right)}{2}\text{ and }\theta \leq \frac{\pi \left(4n_{1}+1\right)}{2}\right)\text{ and }|y|\leq 10\sqrt{\cos(\theta )}$$

求解 y 的值

$$y=\sqrt{100\cos(\theta )-x^{2}}$$

$$y=-\sqrt{100\cos(\theta )-x^{2}}\text{, }\exists n_{1}\in \mathrm{Z}\text{ : }\left(\theta \geq \frac{\pi \left(4n_{1}+3\right)}{2}\text{ and }\theta \leq \frac{\pi \left(4n_{1}+5\right)}{2}\right)\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }\left(\theta \geq \frac{\pi \left(4n_{1}-1\right)}{2}\text{ and }\theta \leq \frac{\pi \left(4n_{1}+1\right)}{2}\right)\text{ and }|x|\leq 10\sqrt{\cos(\theta )}$$