Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
0 votes

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


Can you answer this question?

1 Answer

0 votes
Step 1:
The equations are
$\cos x+\cos y=\large\frac{3}{2}$-----(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)
$\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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App