Finding the Exact X-Coordinates of the Points on the Graph f(x) for Which the Tangent Line is Parallel to the Line g(x) I cannot seem to solve the following situation:
When $f'(x)$ is equal to the slope of $g(x)$ over the interval of $\frac{\pi}{2}\le x \le \pi $ when
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$$f'(x)=\frac{2\cos(2x)}{3\sin(2x)^{\frac{2}{3}} }$$
and $g(x)$ is defined by the equation
$$2x-3(6)^\frac{1}{3}y=0$$
.
So, my fundamental problem is with the trig functions in this case. The problem simplifies to:
$$\frac{\cos(2x)}{\sin(2x)^{\frac{2}{3}}}-\frac{1}{6^\frac{1}{3}}=0$$
.
As always, thanks!

Since $g^\prime(x)=\dfrac{2}{3\sqrt[3]{6}}$ we wish to find a solution for
$$ \frac{2\cos(2x)}{3\sin(2x)^{\frac{2}{3}} }= \dfrac{2}{3\sqrt[3]{6}}$$
in the interval $\frac{\pi}{2}\le x \le \pi$. Simplifying and cubing both sides gives
\begin{eqnarray}
\frac{\cos^3(2x)}{\sin^2(2x)}&=&\frac{1}{6}\\
6\cos^3(2x)-\sin^2(2x)&=&0\\
6\cos^3(2x)+\cos^2(2x)-1&=&0\\
6u^3+u^2-1&=&0\\
\end{eqnarray}
Can you finish from here, given that the polynomial has a rational zero?
