Lahenda sinC=cosB

$$\sin C = \cos B$$

Lahendage ja leidke B (complex solution)

$$\left\{\begin{matrix}B=-i\ln(\frac{\sqrt{2}\left(\sqrt{2}\sin(C)-\sqrt{-\cos(2C)-1}\right)}{2})+2\pi n_{1}\text{, }n_{1}\in \mathrm{Z}\text{, }&\frac{\sqrt{2}\left(\sqrt{2}\sin(C)-\sqrt{-\cos(2C)-1}\right)}{2}\neq 0\\B=-i\ln(\frac{\sqrt{2}\left(\sqrt{2}\sin(C)+\sqrt{-\cos(2C)-1}\right)}{2})+2\pi n_{2}\text{, }n_{2}\in \mathrm{Z}\text{, }&\frac{\sqrt{2}\left(\sqrt{2}\sin(C)+\sqrt{-\cos(2C)-1}\right)}{2}\neq 0\end{matrix}\right,$$

Lahendage ja leidke C (complex solution)

$$\left\{\begin{matrix}C=-i\ln(\frac{2i\cos(B)-\sqrt{2}\sqrt{-\cos(2B)+1}}{2})+2\pi n_{1}\text{, }n_{1}\in \mathrm{Z}\text{, }&\frac{2i\cos(B)-\sqrt{2}\sqrt{-\cos(2B)+1}}{2}\neq 0\\C=-i\ln(\frac{2i\cos(B)+\sqrt{2}\sqrt{-\cos(2B)+1}}{2})+2\pi n_{2}\text{, }n_{2}\in \mathrm{Z}\text{, }&\frac{2i\cos(B)+\sqrt{2}\sqrt{-\cos(2B)+1}}{2}\neq 0\end{matrix}\right,$$

Lahendage ja leidke B

$$B=-\arccos(\sin(C))+2\pi n_{1}\text{, }n_{1}\in \mathrm{Z}$$

$$B=\arccos(\sin(C))+2\pi n_{2}\text{, }n_{2}\in \mathrm{Z}$$

Lahendage ja leidke C

$$C=-\arcsin(\cos(B))+2\pi n_{1}+\pi \text{, }n_{1}\in \mathrm{Z}$$

$$C=\arcsin(\cos(B))+2\pi n_{2}\text{, }n_{2}\in \mathrm{Z}$$