y=1-cos^2x పరిష్కరించండి

$$y = 1 - \cos ^ { 2 } x$$

xని పరిష్కరించండి (సంకీర్ణ పరిష్కారం)

$$x=\left(-i\right)\ln(\left(\left(-1\right)y+1\right)^{\frac{1}{2}}+\left(\left(-1\right)y\right)^{\frac{1}{2}})+2\pi n_{5}\text{, }n_{5}\in \mathrm{Z}$$

$$x=\left(-i\right)\ln(\left(\left(-1\right)y+1\right)^{\frac{1}{2}}+\left(-1\right)\left(\left(-1\right)y\right)^{\frac{1}{2}})+2\pi n_{4}\text{, }n_{4}\in \mathrm{Z}$$

$$x=\left(-i\right)\ln(\left(-1\right)\left(\left(-1\right)y+1\right)^{\frac{1}{2}}+\left(\left(-1\right)y\right)^{\frac{1}{2}})+2\pi n_{17}\text{, }n_{17}\in \mathrm{Z}$$

$$x=\left(-i\right)\ln(\left(-1\right)\left(\left(-1\right)y+1\right)^{\frac{1}{2}}+\left(-1\right)\left(\left(-1\right)y\right)^{\frac{1}{2}})+2\pi n_{16}\text{, }n_{16}\in \mathrm{Z}$$

yని పరిష్కరించండి (సంకీర్ణ పరిష్కారం)

$$y=\frac{-\cos(2x)+1}{2}$$

xని పరిష్కరించండి

$$\left\{\begin{matrix}x=\arccos(\sqrt{1-y})+2\pi n_{1}\text{, }n_{1}\in \mathrm{Z}\text{; }x=-\arccos(\sqrt{1-y})+2\pi n_{2}\text{, }n_{2}\in \mathrm{Z}\text{, }&y\geq 0\text{ and }y\leq 1\text{ and }\sqrt{1-y}\leq 1\\x=-\arccos(\sqrt{1-y})+2\pi n_{3}+\pi \text{, }n_{3}\in \mathrm{Z}\text{; }x=\arccos(\sqrt{1-y})+2\pi n_{4}-\pi \text{, }n_{4}\in \mathrm{Z}\text{, }&y\geq 0\text{ and }y\leq 1\text{ and }-\sqrt{1-y}\geq -1\end{matrix}\right.$$

yని పరిష్కరించండి

$$y=-\left(\cos(x)\right)^{2}+1$$