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Question Number 108547 by 1549442205PVT last updated on 17/Aug/20

$$\mathrm{Solve}\mathrm{the}\mathrm{following}\mathrm{equation}:$$$$\mathrm{cosz}=2$$

Answered by mr W last updated on 17/Aug/20

$$e^{zi}=\cos z+i\sin z$$$$e^{-zi}=\cos z-i\sin z$$$$\Rightarrow \cos z=\frac{e^{zi}+e^{-zi}}{2}=2$$$${\left(e^{zi}\right)}^2-4e^{zi}+1=0$$$$e^{zi}=2\pm \sqrt{3}$$$$zi=\ln (2\pm \sqrt{3})$$$$\Rightarrow z=-i\ln (2\pm \sqrt{3})$$

Commented by 1549442205PVT last updated on 17/Aug/20

$$\mathrm{Thank}\mathrm{Sir}.\mathrm{Result}\mathrm{is}\mathrm{correct}$$$$\mathrm{We}\mathrm{can}\mathrm{also}\mathrm{write}\mathrm{z}=\pm \mathrm{iln}(2+\sqrt{3})$$

Commented by peter frank last updated on 22/Aug/20

$$\mathrm{thank}\mathrm{you}$$

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