Resolver y=lnsqrt{2+cos^2x}

$$y = \ln \sqrt { 2 + \cos ^ { 2 } x }$$

Resolva para x (complex solution)

$$\left\{\begin{matrix}x=-\frac{i\ln(2e^{2y}-2\sqrt{e^{4y}-5e^{2y}+6}-5)}{2}+\pi n_{2}\text{, }n_{2}\in \mathrm{Z}\text{, }&-2e^{2y}+2\sqrt{e^{4y}-5e^{2y}+6}+5\neq 0\text{ and }Im(\ln(e^{y}))-Im(y)=0\\x=-\frac{i\ln(2e^{2y}+2\sqrt{e^{4y}-5e^{2y}+6}-5)}{2}+\pi n_{1}\text{, }n_{1}\in \mathrm{Z}\text{, }&-2e^{2y}-2\sqrt{e^{4y}-5e^{2y}+6}+5\neq 0\text{ and }Im(\ln(e^{y}))-Im(y)=0\end{matrix}\right,$$

Resolva para y (complex solution)

$$y=\ln(\sqrt{2\left(\cos(2x)+5\right)})-\ln(2)$$

$$\nexists n_{2}\in \mathrm{Z}\text{ : }x=\pi n_{2}-\frac{i\ln(5-2\sqrt{6})}{2}+\frac{\pi }{2}\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}-\frac{i\ln(2\sqrt{6}+5)}{2}+\frac{\pi }{2}$$

Resolva para x

$$\left\{\begin{matrix}x=\arccos(\sqrt{e^{2y}-2})+2\pi n_{1}\text{, }n_{1}\in \mathrm{Z}\text{; }x=-\arccos(\sqrt{e^{2y}-2})+2\pi n_{2}\text{, }n_{2}\in \mathrm{Z}\text{, }&y\leq \frac{\ln(3)}{2}\text{ and }y\geq \frac{\ln(2)}{2}\text{ and }\sqrt{e^{2y}-2}\leq 1\\x=-\arccos(\sqrt{e^{2y}-2})+2\pi n_{3}+\pi \text{, }n_{3}\in \mathrm{Z}\text{; }x=\arccos(\sqrt{e^{2y}-2})+2\pi n_{4}-\pi \text{, }n_{4}\in \mathrm{Z}\text{, }&y\leq \frac{\ln(3)}{2}\text{ and }y\geq \frac{\ln(2)}{2}\text{ and }-\sqrt{e^{2y}-2}\geq -1\end{matrix}\right,$$

Resolva para y

$$y=\frac{\ln(\left(\cos(x)\right)^{2}+2)}{2}$$