Question

Need help on this problem

Determine the number(s) between 0 and 2pi where the line tangent to the given function is horizontal

f(x)10sinx-10sqrt(3)cosx

I know I have to take the derivative of it its I just dont know how I solve it for the horizontal line

Answer 1

Do you know to calculate the derivative?

Answer 2

Yes the derivative is
\[f'(x)=10\cos(x)+10\sqrt{3}\sin(x)\]

Answer 3

ok. what is the slope.of horizontal line then?

Answer 4

Zero

Answer 5

So do I set x=0?

Answer 6

well. that's quire typing and i'm on mobile. ill try to help but im slow :(

Answer 7

Thats ok

Answer 8

So I have to do this
\[0=10\cos(0)-10\sqrt{3}\sin(0)\]

Answer 9

Correct?

Answer 10

eh. no. you want to find follow what x makes f'(x) =0

Answer 11

So then
\[0=10\cos(x)+10\sqrt{3}\sin(x)\]

Answer 12

\[-10\cos(x)=10\sqrt{3}\sin(x)\]

Answer 13

Then I divide each one by 10
\[-\cos(x)=\sqrt{3}\sin(x)\]

Answer 14

correct?

Answer 15

so far :) now to the hard part hehe.
I hope I would be ble to help this way..
on worst case ill try when i get home

Answer 16

Ok :D thanks for the guidence

Answer 17

Ok I got the awnsers {(5/6)pi and (11/3)pi}

Answer 18

:D awsome thanks a bunch

Answer 19

cool. :)
ye I just realized how blind I got..
\[cos(x) = -\sqrt{3}sin(x) \\
\frac{cos(x)}{sin(x)} = -\sqrt{3} \\
cot(x) = -\sqrt{3} \]
damn so slow on mobile -.-