wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Solve the following equations
cosxcos2x+cos3x=0.

Open in App
Solution

cosxcos2x+cos3x=0
Use :
cos2x=2cos2x1
cos3x=4cos3x3cosx
Now,
cosxcos2x+cos3x=0
cosx(2cos2x1)+(4cos3x3cosx)=0
4cos3x2cos2x2cosx+1=0
2cos2x[2cosx101[2cosx1]=0
(2cosx1)(2cos2x1)=0
cosx=12=±π3+2πn
cos2x=12cosx=±12
x=π4+2πn.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Basic Theorems in Differentiation
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon