Question Number 10429 by ridwan balatif last updated on 09/Feb/17

2^(cos2x) +2^(cos^2 x) =3×2^(−cos2x)    x=...?  i′m so sorry, it′s my mistake, the true question is  2^(cos2x) +2^(cos^2 x) =3×2^(−cos2π)

Answered by arge last updated on 09/Feb/17

    log2^(cos2x) +log2^(cos^2 x) =log(3×2^(−cos2x) )    log(2^(cos2x) 2^(cos^2 x) )=log3+log2^(−cos2x)         log(2^(cos2x) 2^(cos^2 x) )−log2^(−cos2x) =log3  log[(2^(cos2x) 2^(cos^2 x) )/2^(−cos2x) ]=log3  log[(2^(cos2x) 2^(cos^2 x) )/2^(−cos2x) ]=0.48  2^(cos2x+cos^2 x+cos2x) =3  2^(cos^2 x+2cos2x) =3  cos^2 x+2cos2x=1.58  cos^2 x+2cos^2 x−2sen^2 x=1.58  3cos^2 x−2+2cos^2 x=1.58  5cos^2 x=3.58  cosx=0.85  x=32.20^°

Commented byamir last updated on 09/Feb/17

hello.dear. your method is not correct.  you apply the log in wrong formula.  log(a+b)≠loga×logb

Commented byarge last updated on 10/Feb/17

thank you. Recalls NASA′s mistake.

Answered by sandy_suhendra last updated on 08/Feb/17

2^(cos2x) +2^((cos2x+1)/2) =(3/2^(cos2x) )  2^(cos2x) +(√(2^(cos2x) .2)) = (3/2^(cos2x) )  misalkan 2^(cos2x) =a  a+(√(2a)) = (3/a)  (√(2a)) = (3/a)−a  2a=((3/a)−a)^2   2a=(9/a^2 )−6+a^2   2a^3 =9−6a^2 +a^4   a^4 −2a^3 −6a^2 +9=0  maaf, cuma baru bisa sampai langkah ini

Commented byamir last updated on 09/Feb/17

Commented byamir last updated on 09/Feb/17

a=1.13  2^(cos 2x) =1.13⇒cos 2x=log _2 1.13=.176  2x=cos^(−1) .176=1.39 ⇒x=(1/2)cos^(−1) (.176)+kπ,k∈Z

Answered by amir last updated on 09/Feb/17

2^(cos 2x) =a⇒a+(√(2a))=(3/2)  2(√(2a))=3−2a⇒8a=9−12a+4a^2   4a^2 −20a+9=0⇒a=((20±(√(20^2 −4×4×9)))/(2×4))=  a=((20±16)/8)⇒a=(1/2),(9/2)  2^(cos 2x) =(1/2)⇒cos 2x=−1=cosπ  2x=π+2kπ⇒x=(π/2)+kπ, k∈Z  2^(cos 2x) =(9/2)⇒cos 2x=log_2  9=2log _2 3+1>1.

Commented byridwan balatif last updated on 09/Feb/17

terimakasih bantuannya, setidaknya udah  dapat gambaran penyelesaiannya

Commented byamir last updated on 09/Feb/17

the problem was fixed by sender as   below:  2^(cos 2x) +2^(cos ^2 x) =3×2^(−cos2π)   so: a+(√(2a))=(3/2)  but i solved this case in different post.