Решить y=sqrt{cosx}

$$y = \sqrt { \cos x }$$

Найдите x (комплексное решение)

$$\left\{\begin{matrix}x=\pi n_{1}+\frac{\pi }{2}\text{, }n_{1}\in \mathrm{Z}\text{, }&y=0\\x=-i\ln(-\sqrt{y^{4}-1}+y^{2})+2\pi n_{2}\text{, }n_{2}\in \mathrm{Z}\text{; }x=-i\ln(\sqrt{y^{4}-1}+y^{2})+2\pi n_{3}\text{, }n_{3}\in \mathrm{Z}\text{, }&arg(y)<\pi \text{ and }y\neq 0\end{matrix}\right,$$

Найдите x

$$x=ArcCosI(y^{2})+2\pi n_{7}\text{, }n_{7}\in \mathrm{Z}\text{, }\exists n_{10}\in \mathrm{Z}\text{ : }\left(n_{7}>n_{10}\text{ and }n_{7}<n_{10}+2\right)$$

$$x=2n_{14}\pi +\left(-1\right)ArcCosI(y^{2})\text{, }n_{14}\in \mathrm{Z}\text{, }\exists n_{10}\in \mathrm{Z}\text{ : }\left(n_{10}<n_{14}\text{ and }n_{10}>-2+n_{14}\right)$$

Найдите y

$$y=\sqrt{\cos(x)}$$

$$\exists n_{1}\in \mathrm{Z}\text{ : }\left(x\geq 2\pi n_{1}-\frac{\pi }{2}\text{ and }x\leq 2\pi n_{1}+\frac{\pi }{2}\right)$$