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Question Number 88491 by M±th+et£s last updated on 11/Apr/20

$$\mathit{s}\mathit{o}\mathit{l}\mathit{v}\mathit{e}$$$$cos\left(x\right)=k$$

Answered by mr W last updated on 11/Apr/20

$$if\mid k\mid ⩽1:$$$$\Rightarrow x=2n\pi \pm {\cos }^{-1}k(=realnumber)$$$$$$$$if\mid k\mid >1:$$$$\cos x+i\sin x=e^{ix}$$$$\cos x-i\sin x=e^{-ix}$$$$2\cos x=e^{ix}+e^{-ix}$$$$lett=e^{ix}$$$$2k=t+\frac{1}{t}$$$$t^2-2kt+1=0$$$$t=k\pm \sqrt{k^2-1}$$$$e^{ix}=k\pm \sqrt{k^2-1}$$$$ix=\ln (k\pm \sqrt{k^2-1})$$$$x=-i\ln (k\pm \sqrt{k^2-1})$$$$\Rightarrow x=\pm i\ln (k+\sqrt{k^2-1})(=imaginary)$$

Commented by M±th+et£s last updated on 11/Apr/20

$$thankyousir.buthowcanwegetx=2n\pi \pm {cos}^{-1}\left(k\right)$$

Commented by Kunal12588 last updated on 11/Apr/20

$$genralsolutionof$$$$cosx=m;m\in [-1,1]is$$$$x=2n\pi \pm {cos}^{-1}m;n\in \mathbb{Z}$$

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