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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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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

$(a)\;0\qquad(b)\;imaginary\qquad(c)\;1\qquad(d)\;2$

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1 Answer

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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}$$\cos\large\frac{x-y}{2}=\frac{3}{2}$
$\Rightarrow \cos\large\frac{x-y}{2}=\frac{3}{2}$$>1$
Which has no solution.
$\therefore$ The solution of given equation is imaginary.
answered Oct 17, 2013 by sreemathi.v
 

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