# Math Help - Point on a CSC graph

1. ## Point on a CSC graph

Find the points on the graph of y = -csc(x), where the tangent is parallel to 3y-2x=4. Use exact values.

0 is less than or equal to x which is less than or equal to 2*pi

Thanks

2. Originally Posted by Mr_Green
Find the points on the graph of y = -csc(x), where the tangent is parallel to 3y-2x=4. Use exact values.

0 is less than or equal to x which is less than or equal to 2*pi

Thanks
$3y - 2x = 4\implies y = (2/3)x+(4/3)$.

To be parallel it means the derivative must be equal to 2/3. Now find all point with derivative 2/3.

3. so thh derivate of -csc is =

So I set this equal to 2/3?

4. Originally Posted by Mr_Green
so thh derivate of -csc is =

So I set this equal to 2/3?
Yes.

5. so i get

pi/3

Is this correct? Is this the only point?

6. i found that out by plugging into my calculator and using the intersect ability on the graph. Is there anything mathematical that I could do to solve it?

7. Originally Posted by Mr_Green
Find the points on the graph of y = -csc(x), where the tangent is parallel to 3y-2x=4. Use exact values.

0 is less than or equal to x which is less than or equal to 2*pi

Thanks
Originally Posted by Mr_Green
so i get

pi/3

Is this correct? Is this the only point?
$y = -csc(x) = -\frac{1}{sin(x)}$

$\frac{dy}{dx} = - \left ( - \frac{cos(x)}{sin^2(x)} \right ) = \frac{cos(x)}{sin^2(x)}$

So we need to solve
$\frac{cos(x)}{sin^2(x)} = \frac{2}{3}$

$3cos(x) = 2sin^2(x) = 2(1 - cos^2(x))$

$2cos^2(x) + 3cos(x) - 2 = 0$

Let $z = cos(x)$. Then this equation becomes:
$2z^2 + 3z - 2 = 0$

This has solutions:
$z = -2$ or $z = \frac{1}{2}$

Thus
$cos(x) = -2$ <-- Impossible for real x.
or
$cos(x) = \frac{1}{2} \implies x = \frac{\pi}{3}, \frac{5 \pi}{3}$
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