Vyřešit p^2=cos2θ

$$p ^ { 2 } = \cos 2 \theta$$

Vyřešte pro: p (complex solution)

$$p=-\sqrt{\cos(2\theta )}$$

$$p=\sqrt{\cos(2\theta )}$$

Vyřešte pro: θ (complex solution)

$$\theta =-\frac{i\ln(\sqrt{p^{4}-1}+p^{2})}{2}+\pi n_{1}\text{, }n_{1}\in \mathrm{Z}$$

$$\theta =-\frac{i\ln(-\sqrt{p^{4}-1}+p^{2})}{2}+\pi n_{2}\text{, }n_{2}\in \mathrm{Z}$$

Vyřešte pro: p

$$p=\sqrt{\cos(2\theta )}$$

$$p=-\sqrt{\cos(2\theta )}\text{, }\exists n_{1}\in \mathrm{Z}\text{ : }\left(\theta \geq \pi n_{1}-\frac{\pi }{4}\text{ and }\theta \leq \pi n_{1}+\frac{\pi }{4}\right)$$

Vyřešte pro: θ

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

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