Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
0 votes

Find the general solution for each of the following equation $\cos 3x+\cos x-\cos 2x=0$

$\begin{array}{1 1}(A)\;x=2n\pi\pm \large\frac{\pi}{3}&(B)\;x=n\pi\pm \large\frac{\pi}{3}\\(C)\;x=2n\pi\pm \large\frac{\pi}{4}&(D)\;x=2n\pi\pm \large\frac{\pi}{6}\end{array} $

Can you answer this question?

1 Answer

0 votes
  • $\cos x+\cos y=2\cos \large\frac{x+y}{2}$$\cos \large\frac{x-y}{2}$
  • $\cos x=0\therefore x=(2n+1)\large\frac{\pi}{2}$
  • $\cos x=\cos y$
  • $x=2n\pi\pm y$
$\cos 3x+\cos x-\cos 2x=0$
$2\cos \large\frac{3x+x}{2}$$\cos \large\frac{3x-x}{2}$$-\cos 2x=0$
$2\cos 2x.\cos x-\cos 2x=0$
$\cos 2x(2\cos x-1)=0$
$\therefore$ If $\cos 2x=0,2x=(2n+1)\large\frac{\pi}{2}$
$\Rightarrow x=(2n+1)\large\frac{\pi}{4}$
$\therefore$ If $2\cos x-1=0,\cos x=\large\frac{1}{2}$$=\cos 60^{\large\circ}=\cos \large\frac{\pi}{3}$
$\therefore x=2n\pi\pm \large\frac{\pi}{3}$
Hence (A) is the correct answer.
answered Apr 18, 2014 by sreemathi.v

Related questions

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