Menu Close

cos-x-k-1-k-lt-0-




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

Leave a Reply

Your email address will not be published. Required fields are marked *