x^2-6-1/1-cosθ=secθ का समाधान करें

$$\frac { x ^ { 2 } - 6 - 1 } { 1 - \cos \theta } = \sec \theta$$

x के लिए हल करें (जटिल समाधान)

$$x=-\frac{\sqrt{2}i\left(\cos(\theta )\right)^{-\frac{1}{2}}\sqrt{2\left(-6\cos(\theta )-1\right)}}{2}$$

$$x=\frac{\sqrt{2}i\left(\cos(\theta )\right)^{-\frac{1}{2}}\sqrt{2\left(-6\cos(\theta )-1\right)}}{2}\text{, }\nexists n_{1}\in \mathrm{Z}\text{ : }\theta =2\pi n_{1}\text{ and }\nexists n_{2}\in \mathrm{Z}\text{ : }\theta =\pi n_{2}+\frac{\pi }{2}$$

x के लिए हल करें

$$x=\sqrt{\frac{1}{\cos(\theta )}+6}$$

$$x=-\sqrt{\frac{1}{\cos(\theta )}+6}\text{, }\exists n_{2}\in \mathrm{Z}\text{ : }\left(\left(\theta >\frac{\pi \left(4n_{2}+3\right)}{2}\text{ and }\theta <2\pi \left(n_{2}+1\right)\right)\text{ or }\left(\theta >2\pi n_{2}\text{ and }\theta <\frac{\pi \left(4n_{2}+1\right)}{2}\right)\right)\text{ or }\exists n_{1}\in \mathrm{Z}\text{ : }\left(\theta \geq \frac{4\pi n_{1}+\pi +2\arcsin(\frac{1}{6})}{2}\text{ and }\theta \leq \frac{4\pi n_{1}+3\pi -2\arcsin(\frac{1}{6})}{2}\right)$$

θ के लिए हल करें

$$\theta =-\arccos(\frac{1}{x^{2}-6})+2\pi n_{1}\text{, }n_{1}\in \mathrm{Z}$$

$$\theta =\arccos(\frac{1}{x^{2}-6})+2\pi n_{2}\text{, }n_{2}\in \mathrm{Z}\text{, }|x|\leq \sqrt{5}\text{ or }|x|>\sqrt{7}$$