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
$\displaystyle y = -csc(x) = -\frac{1}{sin(x)}$
$\displaystyle \frac{dy}{dx} = - \left ( - \frac{cos(x)}{sin^2(x)} \right ) = \frac{cos(x)}{sin^2(x)}$
So we need to solve
$\displaystyle \frac{cos(x)}{sin^2(x)} = \frac{2}{3}$
$\displaystyle 3cos(x) = 2sin^2(x) = 2(1 - cos^2(x))$
$\displaystyle 2cos^2(x) + 3cos(x) - 2 = 0$
Let $\displaystyle z = cos(x)$. Then this equation becomes:
$\displaystyle 2z^2 + 3z - 2 = 0$
This has solutions:
$\displaystyle z = -2$ or $\displaystyle z = \frac{1}{2}$
Thus
$\displaystyle cos(x) = -2$ <-- Impossible for real x.
or
$\displaystyle cos(x) = \frac{1}{2} \implies x = \frac{\pi}{3}, \frac{5 \pi}{3}$
So there are two answers.
-Dan