The maximum point on the curve with equation $y = x\sqrt {\sin x} $, $0 < x < \pi $, is the point A, Show that the x-coordinate of point A satisfies the equation $2\tan x + x = 0$
I understand that I need to determine the stationary points on the curve by equating the first differential to 0, however I run into a problem once I find that differential and do that, here's what I've done:
$\eqalign{ & y = x\sqrt {\sin x} = x{\left( {\sin x} \right)^{{1 \over 2}}} \cr & {{dy} \over {dx}} = {1 \over 2}x\cos x{\left( {\sin x} \right)^{ - {1 \over 2}}} + {\left( {\sin x} \right)^{{1 \over 2}}} \cr & {{dy} \over {dx}} = {1 \over 2}{\left( {\sin x} \right)^{ - {1 \over 2}}}\left( {x\cos x + 2\sin x} \right) \cr & 0 = {1 \over 2}{\left( {\sin x} \right)^{ - {1 \over 2}}}\left( {x\cos x + 2\sin x} \right) \cr} $
Where do I go from here? Even if I eliminate the sine expression outside the brackets I still have trouble doing anything with the expression inside the brackets due to the x attached to the cosine. What can be done here? How else can I satisfy this equation?
Thank you!
Hint: rewrite the final expression as $$\frac{x\cos x + 2\sin x}{2(\sin x)^{\frac{1}{2}}} = 0$$.
This is $0$ if and only if $x\cos x + 2\sin x = 0$, or dividing both sides by $\cos x$ (for $x \neq \frac{\pi}{2}$, which is clearly not a solution), we see that this is precisely $$x + 2\tan x = 0$$