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

Solve the following equations.
cos3x2cosx21=132cosx.

Open in App
Solution

cos3x2cosx/21=132cosx
cos2x+cosx21=32cosx [use cosa+cosb=cosa+b2cosab2]
Where a=2xb=x
cos2x2=(31)cosx
2cos2x3=(31)cosx
cosx=t
2t2(31)t3=0
t=(31)±(31)24.2.(3)2.2
t=0.7±4.754
cosx=t=1.063 OR 1.4125
As cosx lies b/w 1 to 1
Hence No solution

1126175_888288_ans_ac6d82a1d2a44de48db7d1612f5e4c07.jpg

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Componendo and Dividendo
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon