求解 x^2-(tanθ+1/tanθ)x+1=0

$$x ^ { 2 } - ( \tan \theta + \frac { 1 } { \tan \theta } ) x + 1 = 0$$

求解 x 的值 (复数求解)

$$x=-\frac{i\left(\sqrt{2\left(-\cos(4\theta )-1\right)}+2i\right)}{2\sin(2\theta )}$$

$$x=-\frac{i\left(-\sqrt{2\left(-\cos(4\theta )-1\right)}+2i\right)}{2\sin(2\theta )}\text{, }\nexists n_{1}\in \mathrm{Z}\text{ : }\theta =\frac{\pi n_{1}}{2}$$

求解 x 的值

$$x=\frac{\sqrt{-\left(\sin(2\theta )\right)^{2}+1}+1}{\sin(2\theta )}$$

$$x=\frac{-\sqrt{-\left(\sin(2\theta )\right)^{2}+1}+1}{\sin(2\theta )}\text{, }\exists n_{1}\in \mathrm{Z}\text{ : }\left(\theta >\frac{\pi n_{1}}{2}\text{ and }\theta <\frac{\pi n_{1}}{2}+\frac{\pi }{2}\right)$$