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)

Question

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

$$2x-3(6)^\frac{1}{3}y=0$$

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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!

Answer 1

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?

Answer 2

Hint: The slope of your $g(x)$ is given by $$\frac{2}{3\cdot 6^{1/3}}$$ and please post your $f(x)$; maybe there is an error in the first derivative. You must prove that this equation $$6\cos(2x)^3+\cos(2x)^2-1=0$$ has no solutions over the given interval.