Problem on finding the maxima of a function I need to find the maximum value of $f(x) = (2\sin A + 3\cos A + 4)^2 * (6-2\sin A-3\cos A)^3$ as $A$ ranges over all real numbers. 
For this I performed the derivative tests by plugging in $2\sin A + 3\cos A = t$, which yielded me the value of $A = \tan^{-1}[-2/3]$. 
Here's where I am stuck. As A lies in the second or fourth quadrant, I am not sure as to what signs I should pick for $\sin A$ and $\cos A$. Calculation gave me the maximum value to be $3456$ at $\sin A = \sqrt{9/13}$
and $\cos A = \sqrt{4/13}$, after taking the positive value of sinA and the negative value of cosA... but why?
Please help me out with this problem. Thanks in advance.

For the ease of calculation, let $t=6-(2\sin A+3\cos A)$
$$\implies6-\sqrt{2^2+3^2}\le t<6+\sqrt{2^2+3^2}$$
$$g(t)=(10-t)^2\cdot t^3$$
Using AM-GM inequality for $(10-t,t>0\iff0<t<10$
$$\dfrac{10}5=\dfrac{2\cdot\dfrac{10-t}2+3\cdot\dfrac t3}{2+3}\ge\sqrt[2+3]{\left(\dfrac{10-t}2\right)^2\left(\dfrac t3\right)^3}$$
the maximum of the RHs will be attained if $$\dfrac{10-t}2=\dfrac t3\iff30-3t=2t\iff t=6$$
$$\iff2\sin A+3\cos A=0$$
$$\iff\dfrac{\sin A}3=\dfrac{\cos A}{-2}=\pm\sqrt{\dfrac{\sin^2A+\cos^2A}{(3)^2+(-2)^2}}$$
So, we need $\sin A=\dfrac{3b}{\sqrt{13}},\cos  A=\dfrac{-2b}{\sqrt{13}}$ where $b=\pm1$
So,  we only need opposite signs of $\sin A,\cos A$
