Given $x, y$ are acute angles such that $\sin y= 3\cos(x+y)\sin x$. Find the maximum value of $\tan y$. 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:
$$(\sin y = 3(\cos x \cos y - \sin x \sin y) \sin x) \frac{1}{\cos y}$$
$$\tan y = (3 \cos x - 3 \sin x \tan y) \sin x$$
$$\tan y + 3 \sin^2 x + \tan y = 3 \sin x \cos x$$
$$\tan y = \frac{3 \sin x \cos x}{1 + 3 \sin^2 x}$$
I have also tried substituting $0$, $30$, $45$, $60$, $90$ to the values of $x$.

Let's play.
$\sin y
= 3\cos(x+y)\sin x
= 3(\cos x \cos y-\sin x \sin y)\sin x
$
Divide by
$\cos x \cos y$
$\dfrac{\tan y}{\cos x}
= 3(\cos x -\sin x \tan y)\tan x
= 3\cos x\tan x -3\sin x\tan x \tan y
$
$\tan y(\dfrac{1}{\cos x}+3\sin x\tan x)
= 3\cos x\tan x
=3\sin x
$
$\tan y(1+3\sin^2 x)
=3\sin x\cos x
$
$\tan y
=\dfrac{3\sin x\cos x}{1+3\sin^2 x}
=\dfrac{\frac32\sin(2 x)}{1+3(1-\cos(2x))/2}
=\dfrac{3\sin(2 x)}{2+3(1-\cos(2x))}
=\dfrac{3\sin(2 x)}{5-3\cos(2x)}
$
At this point,
it being about midnight here,
I threw this at Wolfy
which said that
its max value was
$\dfrac34$
at
$\pi n +\arctan(1/2)$.
So the max value is
$\dfrac34$.
