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Question Number 108547 by 1549442205PVT last updated on 17/Aug/20
$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{following}\:\mathrm{equation}: \\ $$$$\mathrm{cosz}\:=\mathrm{2} \\ $$
Answered by mr W last updated on 17/Aug/20
$${e}^{{zi}} =\mathrm{cos}\:{z}+{i}\:\mathrm{sin}\:{z} \\ $$$${e}^{−{zi}} =\mathrm{cos}\:{z}−{i}\:\mathrm{sin}\:{z} \\ $$$$\Rightarrow\mathrm{cos}\:{z}=\frac{{e}^{{zi}} +{e}^{−{zi}} }{\mathrm{2}}=\mathrm{2} \\ $$$$\left({e}^{{zi}} \right)^{\mathrm{2}} −\mathrm{4}{e}^{{zi}} +\mathrm{1}=\mathrm{0} \\ $$$${e}^{{zi}} =\mathrm{2}\pm\sqrt{\mathrm{3}} \\ $$$${zi}=\mathrm{ln}\:\left(\mathrm{2}\pm\sqrt{\mathrm{3}}\right) \\ $$$$\Rightarrow{z}=−{i}\:\mathrm{ln}\:\left(\mathrm{2}\pm\sqrt{\mathrm{3}}\right) \\ $$
Commented by 1549442205PVT last updated on 17/Aug/20
$$\mathrm{Thank}\:\mathrm{Sir}.\mathrm{Result}\:\mathrm{is}\:\mathrm{correct}\: \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{also}\:\mathrm{write}\:\mathrm{z}=\pm\mathrm{iln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right) \\ $$
Commented by peter frank last updated on 22/Aug/20
$$\mathrm{thank}\:\mathrm{you} \\ $$