G goodnosh7 New member Joined Nov 20, 2007 Messages 2 Nov 20, 2007 #1 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.
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 Elite Member Joined Jan 29, 2005 Messages 11,978 Nov 20, 2007 #2 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.
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.