{l}{cosx=siny}{sin^2y-cosx=2} କୁ ସମାଧାନ କରନ୍ତୁ

$$\left\{ \begin{array} { l } { \cos x = \sin y } \\ { \sin ^ { 2 } y - \cos x = 2 } \end{array} \right.$$

x, y ପାଇଁ ସମାଧାନ କରନ୍ତୁ (ଜଟଳି ସମାଧାନ)

$$x=\left(-i\right)\ln(2+\left(-1\right)\times 3^{\frac{1}{2}})+2\pi n_{1}\text{, }n_{1}\in \mathrm{Z}\text{, }y=\left(-i\right)\ln(2i+\left(-i\right)\times 3^{\frac{1}{2}})+2\pi n_{2}\text{, }n_{2}\in \mathrm{Z}$$

$$x=\left(-i\right)\ln(2+\left(-1\right)\times 3^{\frac{1}{2}})+2\pi n_{1}\text{, }n_{1}\in \mathrm{Z}\text{, }y=\left(-i\right)\ln(2i+i\times 3^{\frac{1}{2}})+2n_{4}\pi \text{, }n_{4}\in \mathrm{Z}$$

$$x=\left(-i\right)\ln(2+3^{\frac{1}{2}})+2\pi n_{5}\text{, }n_{5}\in \mathrm{Z}\text{, }y=\left(-i\right)\ln(2i+\left(-i\right)\times 3^{\frac{1}{2}})+2\pi n_{2}\text{, }n_{2}\in \mathrm{Z}$$

$$x=\left(-i\right)\ln(2+3^{\frac{1}{2}})+2\pi n_{5}\text{, }n_{5}\in \mathrm{Z}\text{, }y=\left(-i\right)\ln(2i+i\times 3^{\frac{1}{2}})+2n_{4}\pi \text{, }n_{4}\in \mathrm{Z}$$

$$x=\pi +2\pi n_{21}\text{, }n_{21}\in \mathrm{Z}\text{, }y=\frac{3}{2}\pi +2\pi n_{22}\text{, }n_{22}\in \mathrm{Z}$$

x, y ପାଇଁ ସମାଧାନ କରନ୍ତୁ

$$x=2\pi n_{1}+\pi$$

$$n_{1}\in \mathrm{Z}$$

$$y=2\pi n_{2}+\frac{3\pi }{2}$$

$$n_{2}\in \mathrm{Z}$$