Given $x, y$ are acute angles such that $\sin y= 3\cos(x+y)\sin x$. Find the maximum value of $\tan y$.
Attempt at a solution:
$[sin y = 3(cos x cos y - sin x sin y) sin x] 1/cos y$
$tan y = (3 cos x - 3 sin x tan y) sin x$
$tan y + 3 sin^2 x + tan y = 3 sin x cos x$
$tan y = (3 sin x cos x) / (1 + 3 sin x)$
I'm stuck here...
Divide both sides by $\cos y$
$$\tan y=3\sin x(\cos x-\sin x\tan y)$$
Divide both sides by $\cos^2x$
$$3\tan x-(3\tan y)\tan^2x=\tan y(1+\tan^2x)$$
$$\iff\tan^2x(4\tan y)-3\tan x+\tan y=0$$
As $\tan x$ is real, the discriminant must be $\ge0$