solve-cos-x-k-

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

$$\boldsymbol{{solve}} \\ $$$${cos}\left({x}\right)={k} \\ $$

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

if ∣k∣≤1: ⇒x=2nπ±cos^(−1) k (=real number) if ∣k∣>1: cos x+i sin x=e^(ix) cos x−i sin x=e^(−ix) 2 cos x=e^(ix) +e^(−ix) let t=e^(ix) 2k=t+(1/t) t^2 −2kt+1=0 t=k±(√(k^2 −1)) e^(ix) =k±(√(k^2 −1)) ix=ln (k±(√(k^2 −1))) x=−i ln (k±(√(k^2 −1))) ⇒x=±i ln (k+(√(k^2 −1))) (=imaginary)

$${if}\:\mid{k}\mid\leqslant\mathrm{1}: \\ $$$$\Rightarrow{x}=\mathrm{2}{n}\pi\pm\mathrm{cos}^{−\mathrm{1}} {k}\:\left(={real}\:{number}\right) \\ $$$$ \\ $$$${if}\:\mid{k}\mid>\mathrm{1}: \\ $$$$\mathrm{cos}\:{x}+{i}\:\mathrm{sin}\:{x}={e}^{{ix}} \\ $$$$\mathrm{cos}\:{x}−{i}\:\mathrm{sin}\:{x}={e}^{−{ix}} \\ $$$$\mathrm{2}\:\mathrm{cos}\:{x}={e}^{{ix}} +{e}^{−{ix}} \\ $$$${let}\:{t}={e}^{{ix}} \\ $$$$\mathrm{2}{k}={t}+\frac{\mathrm{1}}{{t}} \\ $$$${t}^{\mathrm{2}} −\mathrm{2}{kt}+\mathrm{1}=\mathrm{0} \\ $$$${t}={k}\pm\sqrt{{k}^{\mathrm{2}} −\mathrm{1}} \\ $$$${e}^{{ix}} ={k}\pm\sqrt{{k}^{\mathrm{2}} −\mathrm{1}} \\ $$$${ix}=\mathrm{ln}\:\left({k}\pm\sqrt{{k}^{\mathrm{2}} −\mathrm{1}}\right) \\ $$$${x}=−{i}\:\mathrm{ln}\:\left({k}\pm\sqrt{{k}^{\mathrm{2}} −\mathrm{1}}\right) \\ $$$$\Rightarrow{x}=\pm{i}\:\mathrm{ln}\:\left({k}+\sqrt{{k}^{\mathrm{2}} −\mathrm{1}}\right)\:\left(={imaginary}\right) \\ $$

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

thank you sir. but how can we get x=2nπ±cos^(−1) (k)

$${thank}\:{you}\:{sir}.\:{but}\:{how}\:{can}\:{we}\:{get}\:{x}=\mathrm{2}{n}\pi\pm{cos}^{−\mathrm{1}} \left({k}\right) \\ $$

Commented by Kunal12588 last updated on 11/Apr/20

genral solution of cos x = m ; m∈[−1,1] is x=2nπ±cos^(−1) m ; n∈Z

$${genral}\:{solution}\:{of} \\ $$$${cos}\:{x}\:=\:{m}\:;\:{m}\in\left[−\mathrm{1},\mathrm{1}\right]\:\:\:\:{is}\: \\ $$$${x}=\mathrm{2}{n}\pi\pm{cos}^{−\mathrm{1}} {m}\:;\:\:{n}\in\mathbb{Z} \\ $$

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