# Math Help - solving "X" in trig equation

1. ## solving "X" in trig equation

This is my problem:
(image looks funky on here so url provided)
LaTeX:
http://rogercortesi.com/eqn/tempimagedir/eqn9167.png

My line of thinking is shown but the result makes no sense to me. My result just does not seem correct. Can someone show me my error?

2. ## Re: solving "X" in trig equation

What do you not understand? Because I can't see any mistake ...
Calculate the discriminant and afterwars do the back-substitution.

4. ## Re: solving "X" in trig equation

Originally Posted by Bowlbase
First, you mustn't write $\cos=x$, you should write $\cos{\4\theta}=x$.

One root is $\frac{1}{2}(-1-\sqrt{3})$, and the second root is $\frac{1}{2}(\sqrt{3}-1)$.

Now you wrote:

$2x^2+2x-1=0=\frac{1}{2}(-1-\sqrt{3})$, tell me how it is possible?!

That is wrong!

What you can write is the following:

$2x^2+2x-1=(x-\frac{1}{2}(-1-\sqrt{3}))(x-\frac{1}{2}(\sqrt{3}-1))$

If we return back to origins, we get:

$\cos{4\thete}=\frac{1}{2}(-1-\sqrt{3}) \approx -1.3$

It is impossible!

So we left only with:

$\cos{4\thete}=\frac{1}{2}(\sqrt{3}-1)$

So, why you took the $Arcsin$ from both sides?

5. ## Re: solving "X" in trig equation

gah, don't know why I used arcsin all of a sudden.

let me rewrite this:

http://rogercortesi.com/eqn/tempimagedir/eqn9167.png

I dont know how to do the little +/- in LaTeX so I just split them with division symbol.

The answer is looking for all values between 0-360 degrees. I just didn't type all that out.

6. ## Re: solving "X" in trig equation

Originally Posted by Bowlbase
gah, don't know why I used arcsin all of a sudden.

let me rewrite this:

http://rogercortesi.com/eqn/tempimagedir/eqn9167.png

I dont know how to do the little +/- in LaTeX so I just split them with division symbol.

The answer is looking for all values between 0-360 degrees. I just didn't type all that out.
So, $0=\frac{1}{2}(-1\pm\sqrt{3})$ ?

Is the above looks right to you?!

7. ## Re: solving "X" in trig equation

I didn't mean to put 0= 1/2(-1+/-sqrt[3]). should be x. I just didn't put an x in the next line.

http://rogercortesi.com/eqn/tempimagedir/eqn9167.png