# Trig Derivative Q

• Mar 17th 2010, 04:46 PM
mneox
Trig Derivative Q
Find points on the function f(x) = 2sinx + sin²x for when the tangent line is horizontal.

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so i found the derivative of that function, which is
f'(x) = 2cosx + 2sinxcosx

what do i do now? set f'(x) to 0? i tried that, but the answer key says that the answer is

(π/2 + 2nπ, 3), (3π/2 + 2nπ, -1)

what does that notation mean? i am confused.. thx
• Mar 17th 2010, 05:05 PM
skeeter
Quote:

Originally Posted by mneox
Find points on the function f(x) = 2sinx + sin²x for when the tangent line is horizontal.

---------

so i found the derivative of that function, which is
f'(x) = 2cosx + 2sinxcosx

what do i do now? set f'(x) to 0? i tried that, but the answer key says that the answer is

(π/2 + 2nπ, 3), (3π/2 + 2nπ, -1)

what does that notation mean? i am confused.. thx

$\displaystyle f'(x) = 2\cos{x}(1 + \sin{x})$

the base solutions in the interval $\displaystyle [0, 2\pi]$ are ...

$\displaystyle \cos{x} = 0$ ... $\displaystyle x = \frac{\pi}{2} \, , \, \frac{3\pi}{2}$

$\displaystyle \sin{x} = -1$ ... $\displaystyle x = \frac{3\pi}{2}$

the $\displaystyle + 2n\pi$ includes all the other solutions that are coterminal with the base solutions.
• Mar 17th 2010, 05:18 PM
mneox
Quote:

Originally Posted by skeeter
$\displaystyle f'(x) = 2\cos{x}(1 + \sin{x})$

the base solutions in the interval $\displaystyle [0, 2\pi]$ are ...

$\displaystyle \cos{x} = 0$ ... $\displaystyle x = \frac{\pi}{2} \, , \, \frac{3\pi}{2}$

$\displaystyle \sin{x} = -1$ ... $\displaystyle x = \frac{3\pi}{2}$

the $\displaystyle + 2n\pi$ includes all the other solutions that are coterminal with the base solutions.

thanks, i get that part now, but do you know what the 3 and -1 mean? How did those y-values come about?
• Mar 17th 2010, 05:40 PM
skeeter
Quote:

Originally Posted by mneox
thanks, i get that part now, but do you know what the 3 and -1 mean? How did those y-values come about?

plug the x-value solutions into the original function.