{cos2x > 0, sinx > 0}
Пусть log[cos2x](sinx) = t
тогда 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 = 0
t = 1/2
log[cos2x](sinx) = 1/2
log[sinx](cos2x) = 2
sin^2x = cos2x
sin^2x = cos^2x - sin^2x
2sin^2x = cos^2x | : cos^2x
2tg^2x = 1
tg^2x = 1/2
tgx = +-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