I need help on this

[natHenderson]

I figured that the area would be equal to x*cos(x), so I took the derivative of that which gave me cos(x) - xsin(x), and I set that equal to zero but it gave me nothing.

 

[Harry_the_cat]

Actually the area is 2x*cos(x). But you still end up with cos(x) - xsin(x) =0.
A question first: Can you use a graphics calculator to help with this?

 

[Dr.Peterson]

View attachment 19048
I figured that the area would be equal to x*cos(x), so I took the derivative of that which gave me cos(x) - xsin(x), and I set that equal to zero but it gave me nothing.

Can you show the work that "gave you nothing"? How did you try to solve it? In what sense did you get nothing?

I can tell you that the equation can't be solved exactly, which is what Harry is implying.

 

[HallsofIvy]

Yes, you have $\displaystyle cos(x)= xsin(x)$ so $\displaystyle x= cot(x)$. There is no "algebraic" way to solve that but you could use some numeric method. For example I note that if $\displaystyle x= \pi/4$, $\displaystyle cot(\pi/4)= 1$ which is larger than $\displaystyle \pi/4= 0.785...$. But if $\displaystyle x= \pi/2$, $\displaystyle cot(\pi/2)= 0$ which is smaller than $\displaystyle x= \pi/2$. There must be an x between $\displaystyle \pi/4$ and $\displaystyle \pi/2$ such that $\displaystyle x= cot(x)$. Where? We don't know but we could try, say, half way between: $\displaystyle \frac{\pi/4+ \pi/2}{2}= \frac{3\pi}{8}$. $\displaystyle cot(3\pi/8)= 0.414$ which is smaller than $\displaystyle \frac{3\pi}{8}= 1.178$. So there must be a solution between $\displaystyle \frac{3\pi}{8}$ and $\displaystyle \frac{\pi}{4}$. Again, try half way between, $\displaystyle \frac{\frac{3\pi}{8}+ \frac{\pi}{4}}{2}= \frac{5\pi}{16}$