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Question Number 89592 by M±th+et£s last updated on 18/Apr/20

cos(x)=k   {−1≤k<0}

$${cos}\left({x}\right)={k}\: \\ $$ $$\left\{−\mathrm{1}\leqslant{k}<\mathrm{0}\right\} \\ $$

Commented bymr W last updated on 18/Apr/20

i don′t understand what′s your problem.    if cos (x)=k with −1≤k≤1, then  x=2nπ±cos^(−1) (k) always!

$${i}\:{don}'{t}\:{understand}\:{what}'{s}\:{your}\:{problem}. \\ $$ $$ \\ $$ $${if}\:\mathrm{cos}\:\left({x}\right)={k}\:{with}\:−\mathrm{1}\leqslant{k}\leqslant\mathrm{1},\:{then} \\ $$ $${x}=\mathrm{2}{n}\pi\pm\mathrm{cos}^{−\mathrm{1}} \left({k}\right)\:{always}! \\ $$

Commented byM±th+et£s last updated on 18/Apr/20

i thought that for k>0

$${i}\:{thought}\:{that}\:{for}\:{k}>\mathrm{0} \\ $$

Answered by ajfour last updated on 18/Apr/20

or x=(2n+1)π±cos^(−1) (−k)

$${or}\:{x}=\left(\mathrm{2}{n}+\mathrm{1}\right)\pi\pm\mathrm{cos}^{−\mathrm{1}} \left(−{k}\right) \\ $$

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