# Math Help - Need help on derivatives of trignometric function

1. ## Need help on derivatives of trignometric function

Find all points on the curve below at which the tangent line is horizontal. y = (cos x) / (2 + sin x)

can somebody help me to solve this question?

2. Originally Posted by btwest69
Find all points on the curve below at which the tangent line is horizontal. y = (cos x) / (2 + sin x)

can somebody help me to solve this question?

Hint: the tangent line is horizontal if it has a slope of zero, that is, the derivative is zero. so find the derivative, set it equal to zero and solve for the x's

3. Originally Posted by Jhevon
Hint: the tangent line is horizontal if it has a slope of zero, that is, the derivative is zero. so find the derivative, set it equal to zero and solve for the x's

-2sinx -1 / (2+sinx)^2
I am not sure of my answer for derivative

4. Originally Posted by btwest69
-2sinx -1 / (2+sinx)^2
I am not sure of my answer for derivative
yes it is, if by that you mean $\frac {-2 \sin x - 1}{(2 + \sin x )^2}$

now set that to zero, and solve for x

5. Originally Posted by Jhevon
yes it is, if by that you mean $\frac {-2 \sin x - 1}{(2 + \sin x )^2}$

now set that to zero, and solve for x

i am stuck with this step
can you show me how to make it to 0

6. Originally Posted by btwest69
i am stuck with this step
can you show me how to make it to 0
a fraction is zero when its numerator is zero (provided its denominator is not zero at the same time). just set the numerator equal to zero and solve for x

7. Originally Posted by Jhevon
a fraction is zero when its numerator is zero (provided its denominator is not zero at the same time). just set the numerator equal to zero and solve for x

-2 \sin x - 1}{(2 + \sin x )^2=0

my answer for x are 2 and 1/2

8. Originally Posted by btwest69

-2 \sin x - 1}{(2 + \sin x )^2=0

my answer for x are 2 and 1/2

what?

$-2 \sin 2 - 1 \ne 0$

and $-2 \sin \frac 12 - 1 \ne 0$

you have a trig equation here. the x's will be angles, and you will have infinitely many answers, not just 2

as i said, you need to solve $-2 \sin x - 1 = 0 \implies \sin x = - \frac 12$

what $x$'s does that happen for?

EDIT: I'm going to bed, so i won't be able to answer you if you respond, so i'll just give you the answers.

$x = \frac {7 \pi}6 + 2n \pi$ for $n \in \mathbb{Z}$ or $x = \frac {11 \pi}6 + 2k \pi$ for $k \in \mathbb{Z}$