Алгебра, логарифмические уравнения и неравенства. Пожалуйста, помогите решить

a)
3^((log(2) x)^2 - log(2) x) = (1\27)^(log(2) 1\x)
ОДЗ: x > 0
__ 1\27 = 1\3^3 = 3^(-3)
__ (1\27)^(log(2) 1\x = [3^(-3)]^((log(2) 1\x) = 3^[-3*log(2) 1\x] =
=>
3^ ((log(2) x)^2 - log(2) x) = 3^ [(-3)*log(2) 1\x]
(log( - log(2) x) = [(-3)*(-1)*log(2) x]
log(2) x)^2 - log(2) x = 3*log(2) x
log(2) x)^2 - 4*log(2) x = 0
log(2) x * (log(2) x - 4) = 0
a) log(2) x = 0 -----------------------------------> x = 2^0 = 1
б) (log(2) x - 4) = 0 -----> log2(x) = 4 -----> x = 2^4 = 16
б)
3^ ((log(2) x)^2 - log(2) x^3) = (1\9)^ (log(2) 1\x + 3)
3^ ((log(2) x)^2 - log(2) x^3) = ((3^(-2))^ (log(2) 1\x + 3)
3^ ((log(2) x)^2 - log(2) x^3) = 3^ [(-2)*(log(2) 1\x + 3]
((log(2) x)^2 - 3*log(2) x^3) = [(-2)*(log(2) 1\x + 3]
((log(2) x)^2 - 3*log(2) x^3) = [(-2)(-1)*(log(2) x - 6]
(log(2) x)^2 - 3*log(2) x^3 = 2*log(2) x - 6
(log(2) x)^2 - 5*log(2) x^3 + 6 = 0 ------> log(2) x = t
t^2 - 5t + 6 = 0 ------> t1 = 2; t2 = 3
t1 = 2 ----> log(2) x = 2 -----> x = 2^2 = 4
t2 = 3 ----> log(2) x = 3 -----> x = 2^3 = 8
в)
[log(V2) cos x + 1] \ (2x^2 + 0,3) > 0
а)
{ log(V2) cos x + 1 < 0
{ 2x^2 + 0,3 < 0
=>
{ log(V2) cos x < 1
{ 2x^2 < - 0,3 -- быть не может
б)
{ log(V2) cos x + 1 > 0
{ 2x^2 + 0,3 > 0
=>
{ log(V2) cos x > -1 -----> log(V2) cos x > log(V2) (V2)^(-1)
{ x^2 > - 0,15 -----> x - любое
=>
cos x > 1\V2
x > pi\4 + 2pi*n

=Ю