=>

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.

]]>2sinx.cosx + cos^2x = 1

]]>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.]]>

x^3 + 2x^2 - 9x = 18

x^3 + 2x^2 - 9x-18 = 0

x^2(x+2)-9(x+2)=0

(x+2)(x^2-9)=0

x=-2, 3

Check: -2

-2^3+2(-2)^2-9(-2)=18

-8+8+18=18

18=18

Check: 3

3^3 + 2(3)^2 - 9(3) = 18

27+18-27=18

18=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.

]]>Solve for X:

x^3+2x^2-9x=18

x^3+2x^2-x=-2 Is this right? If it is then, I can't seem to go any further

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).

]]>x^(3/4) + 5/x^(1/4) (you may assume x>0)

And how do I find the absolute inequality corresponding to: x<=-22/3 OR x>=11/3 (And what exactly is the absolute inequality?)

]]>