Hi folks,

most problems can be done in more than one way, but all should give the same answer.

In the following, I get 2 coordinates using method 1 and 3 coordinates when using method 2.

for the curve $y = cos^3 x. sin x$ where $0 \le x \le \pi$ find the coordinates of the points where $\dfrac{dy}{dx} = 0$

so, method 1

$\dfrac{dy}{dx} = - 3 cos^2 x. sin^2 x + cos^3 x. cos x = 0$

$ y' = cos^4 x - 3 cos^2 x. sin^2 x = 0$

$ y' = cos^2 x ( cos^2 x - 3 sin^2 x) = 0$

so $cos x = 0$

or

$cos^2 x = 3 sin^2 x \Rightarrow tan^2 x = \dfrac{1}{3} \Rightarrow tan x = \dfrac{1}{\sqrt{3}}$

gives $x = cos^{-1}(0) \Rightarrow x = \dfrac{\pi}{2} and \dfrac{3\pi}{2}$ but only $\dfrac{\pi}{2}$ is in quad 1 and 2, so $x = \dfrac{\pi}{2}$ is the only solution.

OR $x = tan^{-1} (\dfrac{1}{\sqrt{3}}) \Rightarrow x = \dfrac{\pi}{6}$ and $\dfrac{7\pi}{6}$ but range is only in quad 1 and 2 so solution in quad 3 is out of range.

There is therefore only one solution which is $\dfrac{\pi}{6}$

therefore the coordinates where $y' = 0$ are $(\dfrac{\pi}{2}, 0)$ and $(\dfrac{\pi}{6}, \dfrac{3\sqrt{3}}{16})$

but, method 2

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

I use $cos^2 x \equiv \frac{1}{2}(1 + cos 2x)$ and $sin^2 x \equiv 1 - cos^2 x$ and get:

$cos 2x = - 1$ or $cos 2x = \frac{1}{2}$

and this gives a third coordinate at $(\dfrac{5\pi}{6}, - \dfrac{3\sqrt{3}}{16})$ which is the right answer.

so what is wrong with method 1?