PLEASE HELP ME!!!!.... easy calculus

[goodnosh7]

Given the maximum value of [1/(4sin(theta) + 3cos(theta) + k)] is 2, for theta is between 0 and 360 degrees. find the value of k.

 

[pka]

The maximum occurs if the derivative is zero.
$\displaystyle - \left( {4\cos (\Theta ) - 3\sin (\Theta )} \right) = 0\; \Rightarrow \;\Theta = \arctan \left( {\frac{4}{3}} \right) \vee \arctan \left( {\frac{4}{3}} \right) + 180^ \circ$.

Now you must solve this for k.
$\displaystyle \frac{1}{{4\sin (\Theta ) + 3\cos (\Theta ) + k}} = 2$

I get two answers.