1. ## Please solve this equation for me!

I know this isn't very hard, but I seem to keep on messing up somewhere

Using an extra variable A for simplicity,
A= 8cosX *(sqrt(10) / (sqrt(10)+2)

Solve for cosX (or (cosX)^2, whichever is found easier) in:
5A/(40-A^2)=cosX

2. Originally Posted by machack
I know this isn't very hard, but I seem to keep on messing up somewhere

Using an extra variable A for simplicity,
A= 8cosX *(sqrt(10) / (sqrt(10)+2)

Solve for cosX (or (cosX)^2, whichever is found easier) in:
5A/(40-A^2)=cosX

$A= 8\cos x \times \frac{\sqrt{10}}{\sqrt{10}+2}$

then

$\frac{A}{ \frac{8\sqrt{10}}{\sqrt{10}+2}}= \cos x$

$\cos x=\frac{A}{ \frac{8\sqrt{10}}{\sqrt{10}+2} }$

$\cos x=\frac{A(\sqrt{10}+2)}{8\sqrt{10}}$

3. No thats not the equation. A was an arbitrary variable made by me to make the real equation (the 2nd one) more managable to type. You gotta plug in A into the 2nd equation and solve for cos x.

4. Originally Posted by machack
I know this isn't very hard, but I seem to keep on messing up somewhere

Using an extra variable A for simplicity,
A= 8cosX *(sqrt(10) / (sqrt(10)+2)

Solve for cosX (or (cosX)^2, whichever is found easier) in:
5A/(40-A^2)=cosX

Ok then,

$\frac{A}{(40-A^2)}=\cos x$

$A= \frac{8\sqrt{10}\cos x}{\sqrt{10}+2}$

$\frac{\frac{8\sqrt{10}\cos x}{\sqrt{10}+2}}{(40-(\frac{8\sqrt{10}\cos x}{\sqrt{10}+2})^2)}=\cos x$

$\frac{\frac{8\sqrt{10}\cos x}{\sqrt{10}+2}}{(40-(\frac{8\sqrt{10}\cos x}{\sqrt{10}+2})^2)}=\cos x$

$\frac{\frac{8\sqrt{10}\cos x}{\sqrt{10}+2}}{(40-(\frac{640\cos^2 x}{(\sqrt{10}+2)^2}))}=\cos x$

You then need to make a common denominator in the 'overall' denominator.

Then divide out the fraction by multiplying the reciprocal, you should be able to then cancel out some of the $\cos x$

Post what you try. I'll have a look.

5. Ah i finally managed to solve this. It turned out your helped since solving for A first then plugging in the value of A to solve for cos^2(x) turned out to be simpler than solving for cos^2(x) right away.