Question

Find the area of the region enclosed between y=4sin(x) and y=2cos(x) from x=0 to x=0.8pi.

Answer 1

these curves cross where
\[4\sin(x)=2\cos(x)\] and it is not clear to me how you are supposed to find this point

Answer 2

I know the area needs to be calculated in two parts. But I don't know where they cross so I can tell which function is on top and which is on bottom.

Answer 3

http://www.wolframalpha.com/input/?i=y%3D4sin%28x%29%2Cy%3D2cos%28x%29
first cosine is greater, then sine is

Answer 4

that you can see from the picture. where they cross is a different story
http://www.wolframalpha.com/input/?i=4sin%28x%29%3D2cos%28x%29

Answer 5

That's the part I'm trying to figure out.

Answer 6

well good luck. apparently the answer is
\[\tan^{-1}(2+\sqrt{5})\]

Answer 7

the .8pi suggests that maybe you are supposed to be using a calculator or algebra software

Answer 8

i.e. doing it numerically with a machine of some kind. is that the case?