Question Number 196517 by ERLY last updated on 26/Aug/23
Answered by mr W last updated on 27/Aug/23
$$\mathrm{cos}\:{x}+{i}\:\mathrm{sin}\:{x}={e}^{{ix}} \\ $$$$\mathrm{cos}\:{x}−{i}\:\mathrm{sin}\:{x}={e}^{−{ix}} \\ $$$$\Rightarrow\mathrm{cos}\:{x}=\frac{{e}^{{ix}} +{e}^{−{ix}} }{\mathrm{2}}=\mathrm{2} \\ $$$$\Rightarrow\left({e}^{{ix}} \right)^{\mathrm{2}} −\mathrm{4}\left({e}^{{ix}} \right)+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{e}^{{ix}} =\mathrm{2}\pm\sqrt{\mathrm{3}} \\ $$$$\Rightarrow{ix}=\mathrm{ln}\:\left(\mathrm{2}\pm\sqrt{\mathrm{3}}\right)=\pm\mathrm{ln}\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right) \\ $$$$\Rightarrow{x}=\mathrm{2}{k}\pi\pm{i}\:\mathrm{ln}\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\:\checkmark \\ $$