Can you do this??????????????

Question

2cos^2 x-(2+3^(1/2))cos x+3=0
find x

Answer 1

Howdy! :)

I sure hope you're doing well. :) As for the question:

2Cos^2x -- (2 + 3^1/2) CosX + 3 = 0 Cannot be factored!

Therefore, to solve, we must use the formula:

aY^2 + bY + c = 0, Wherein:

(+-- means + or --!)

Y = [-- b +-- (b2 -- 4ac)^1/2]/2a

In this case, Y = CosX, a =2, b = -- (2 + 3^1/2) & c = 3 So:

CosX = { (2 + 3^1/2) +-- [(2 + 3^1/2)^2 -- 4x3x2] }/(2x2)

But, -- 4x3x2 = -- 24 is greater than (2 + 3^1/2)^2 which (approximately) = -- 13.928!

& it is "illegal" to take the square root of a negative number!

So, we must complete the square:

First, we divide both sides of the equation by 2. We get:

CosX^2 -- {[2 + (3^1/2)]/2}CosX + 3/2 = 0

Next, we subtract 3/2 from both sides of the equation:

CosX^2 -- {[2 + (3^1/2)]/2}CosX = -- 3/2

Now, to simplify matters, I will use the approximate measure of {[2 + (3^1/2)]/2} = (approximately) 1.866. Leaving it as {[2 + (3^1/2)]/2} works too, but is far more cumborsome, especially as you end up having quite a few terms, which increases the possibiliy of making an error!

Now, we add the square of 1/2 of the coefficient of CosX. Meaning, we add 0.8705 to both sides of the equation:

CosX^2 -- 1.866CosX + 0.8705 = 0.8705 -- 3/2

CosX^2 -- 1.866CosX + 0.8705 = -- 0.62948

At this stage, we're supposed to take the square root of both sides. But, it is "illegal" in math to take the square root of a negative number!

So, CosX is undefined, since CosX is only defined over the interval [--1, 1]!

([--1, 1] means the interval includes --1 & 1, as well as all the numbers in between. [--1, 1) means the interval includes --1 & all the numbers between --1 & 1, but excludes 1. Whereas (--1, 1] includes 1 & all the numbers between --1 & 1, but excludes --1!)

If this question is a Calculus question, then this is where it ends. But if it's an algebra question, then you can use imaginary numbers (i^2 = --1) to solve for CosX and then X.

I sure hope that helps and is as clear as can be for you to easily understand how to solve the question. :) Good luck, take very good care and have a great day & a great weekend. :)

Cheers! :)

Answer 2

7

Answer 3

1) This is just a standard quadratic: a(cosx)^2 + b(cosx) +c = 0; where cosx is the unknown, a = 2, b = 2 + SQRT(3), c = 3.

2) Just solve for cosx using the quadratic formula. cosx = (-b +/-SQRT(b^2 - 4ac)) / 2a

3) There will be two solutions for cosx; call them x1 and x2. The two solutions correspond to the +/- preceeding SQRT in the formula.

4) The x you want to find, is, the x, whose cos is x1 or x2; usually called the arccos. If x is unconstrained there can be many solutions; but, if x is in the open interval [0, PI) there will be no more than one.

Answer 4

I can, but I won't. Do your homework at home. Make sure your calulator is in degrees mode for the cosine part.

Answer 5

The equation has no solution in the real number domain.....

Answer 6

wow 2=pi...
i think

Answer 7

no I can't

Answer 8

omg-no!

Answer 9

Not "no"...but "hell no"!!