Multiply the first term by x^(1/4)/x^(1/4) :

x^(3/4) * x^(1/4)/x^(1/4) + 5/x^(1/4) 

= x^((3/4)+(1/4))/x^(1/4) + 5/x^(1/4) 

= x^1/x^(1/4) + 5/x^(1/4) 

= x / x^(1/4) + 5/x^(1/4)

= (x+5) / x^(1/4)

Absolute Inequality I think refers to something like |x|<3, therefore -3<x<3, so we need an absolute value (in other words it is the distance from 0, ignoring plus or minus)

x<=-22/3 OR x>=11/3 could possibly be solved by taking the difference of the two absolute values:

|-22/3| - |11/3| = 11/3

Then halve that and put it in a formula: |x+11/6|<11/3+11/6  ==> |x+11/6|<11/2

Why did I do this? Because I wanted to know how to find the "middle" of those two numbers. (Try putting dots on the number line to see what I mean).

x³ + 2x² - 9x = 18

x³ + 2x² - 9x - 18 = 0
Putting x=3, we find that it is one of the solutions.
Therefore,
(x-3)(x² + 5x + 6) = 0
x² + 5x + 6 = (x+3)(x+2)
Therfore,
(x-3)(x+3)(x+2) = 0
which gives the following values of x :- 3, -3, -2

PS:_ A cubic equation, that is an equation of degree three, can be solved only by trial and error, as far as I know.

What you've done is correct, just incomplete.
For the x²-9=0 bit, that means that x²=9, so x=√9.
Square roots always have two answers, positive and negative, so x=±3.
That means that your solutions are -2, -3 and 3, which agrees with ganesh's workings.

2sinxcosx + cos^2x = 1

=>

2sinxcosx - sin^2x = 0 (bc of the pythagorean identity)

=>

(2cosx - sinx)sinx = 0

=>

2cosx=sinx or sinx = 0

=>

tan x = 2 or sin x = 0

so x = arctan 2 or x = n(pi), where n is any integer.