# The solution set of the system of equational $x+y=\large\frac{2\pi}{3}$$,\cos x+\cos y=\large\frac{3}{2} where x and y are real is (a)\;0\qquad(b)\;imaginary\qquad(c)\;1\qquad(d)\;2 ## 1 Answer Step 1: The equations are x+y=\large\frac{2\pi}{3}-----(1) \cos x+\cos y=\large\frac{3}{2}-----(2) \Rightarrow \large\frac{2}{\cos x}=\frac{\cos(x+y)+\cos(x-y)}{\cos(x+y)\cos(x-y)} \Rightarrow \large\frac{2}{\cos x}=\frac{2\cos x-\cos y}{\cos^2x-\sin^2y} \cos^2x-\sin^2y=\cos^2 x\cos y \Rightarrow \cos^2x(1-\cos y)=\sin ^2y \Rightarrow \cos^2x=\large\frac{1-\cos^2y}{1-\cos y} \qquad\quad\;\;\;=1+\cos y \Rightarrow \cos^2x=2\cos^2\big(\large\frac{y}{2}\big) \Rightarrow \cos^2x\sec^2\large\frac{y}{2}$$=2$
$\Rightarrow \cos x\sec\big(\large\frac{y}{2}\big)$$=\pm \sqrt 2 Step 2: From equation (2) 2\cos\large\frac{x+y}{2}$$\cos\large\frac{x-y}{2}=\frac{3}{2}$
$\Rightarrow 2\cos\large\frac{\pi}{3}$$\cos\large\frac{x-y}{2}=\frac{3}{2}[Using equation (1)] \Rightarrow 2.\large\frac{1}{2}$$\cos\large\frac{x-y}{2}=\frac{3}{2}$