InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
The number of solution of `sin x+sin 2x+sin 3x` `=cos x +cos 2x+cos 3x, 0 le x le 2pi`, isA. 7B. 5C. 4D. 6 |
|
Answer» Correct Answer - D We have `(sin x+ sin 3x)+sin 2x=(cos x+cos 3x)+cos 2x` or `2 sin 2x cos x+ sin 2x=2 cos 2x cos x + cos 2x` or `sin 2x(2 cos x +1)=cos2x(2 cso x +1)` or `(2 cos x+1) (sin 2x- cos 2x)=0` or `cos x=-1//2 or sin 2x-cos 2x=0` `rArr x=2 n pi pm (2 pi//3) or tan 2x=1=tan (pi//4)` `rArr x=2n pi pm (2pi//3) or x=(4n+1) pi//8, n in Z` But here `0 le x le 2pi`. Hence, `x=pi//8, 5pi//8, 2pi//3, 9 pi//8, 4pi//3, 13 pi//8`. |
|