Пожалуйста помогите с решением[tex]4log_cos2xsinx-4+3log_sin^3xcos2x=0 [/tex]

cos2x gt; 0, sinx gt; 0
Пусть log[cos2x](sinx) = t
тогда log[sin^3x](cos2x) = 1/log[cos2x](sin^3x) = 1/(3log[cos2x](sinx)) = 1/(3t)

--gt;
4t - 4 + 1/t = 0 t lt;gt; 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)
--gt;
sinx = 1/sqrt(3) sinx gt; 0
cosx = +-sqrt(2)/sqrt(3) -gt; cos2x = 2/3 - 1/3 = 1/3
--gt;
x = arctg(1/sqrt(2)) + 2Пk, x = П - arctg(1/sqrt(2)) + 2Пk