Find equations of the two lines through the origin that make an angle $\tan^{-1}(1/2)$ with $3y=2x$ 
If two straight lines pass through the origin and makes an angle $\tan^{-1}(1/2)$ with $3y=2x$, then find its equations.

Let $m$ be the gradient of the line then, $$\frac{1}{2}=\frac{m-2/3}{1-2m/3}$$ I don't know whether my approach is correct or not. 

$tan^{-1}(\frac{1}{2}) = \theta$
$tan(\theta) = \frac{1}{2}$
Line $3y = 2x$ have the slope of $\frac{2}{3}$, suppose this line is called a and the two lines in question is b and c (with $m_{b} > m_{a}, m_{c} < m_{a})$.
So $m_{a} = \frac{2}{3}$ $=>$ $tan(A) = \frac{2}{3}$, $m_{b} = tan(B)$, and $m_{c} = tan{C}$.
Hint :
$B - A = tan^{-1}(\frac{1}{2})$
$A-C = tan^{-1}(\frac{1}{2}) $
$tan(B-A) = \frac{1}{2}$ 
$tan(A-C) = \frac{1}{2}$
$tan(B)$ and $tan(C)$ will be the slope of the line you want to find, with $tan(A) = \frac{2}{3}$.
Rather complicated but i think the easiest one you can do with logical thinking rather than looking up formula
