Пожалуйста помогите с решением
[tex]4log_{cos2x}sinx-4+3log_{sin^3x}cos2x=0 [/tex]

большое спасибо!

{cos2x > 0, sinx > 0}Пусть log[cos2x](sinx) = tтогда log[sin^3x](cos2x) = 1/log[cos2x](sin^3x) = 1/(3log[cos2x](sinx)) = 1/(3t)--> 4t - 4 + 1/t = 0 {t <> 0}4t^2 - 4t + 1 = 0(2t - 1)^2 = 0t = 1/2log[cos2x](sinx) = 1/2log[sinx](cos2x) = 2sin^2x = cos2xsin^2x = cos^2x - sin^2x2sin^2x = cos^2x | : cos^2x2tg^2x = 1tg^2x = 1/2tgx = +-1/sqrt(2)-->sinx = 1/sqrt(3) {sinx > 0}cosx = +-sqrt(2)/sqrt(3) -> cos2x = 2/3 - 1/3 = 1/3-->x = arctg(1/sqrt(2)) + 2Пk, x = П - arctg(1/sqrt(2)) + 2Пk