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

Solve : secx1=(21)tanx,x(2n1)π2,nZ

A
2nπ,(4n+1)π2,nZ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
nπ,(4n+1)π2,nZ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2nπ,2nπ+π4,nZ
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
2nπ,(2n+1)π2,nZ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 2nπ,2nπ+π4,nZ
secx1=(21)tanx
1cosx1=tanπ8tanx

1cosxcosx=sinxsinπ8cosxcosπ8

cosπ8cosxcosπ8=sinxsinπ8

sinxsinπ8+cosxcosπ8=cosπ8

cos(xπ8)=cosπ8

xπ8=2nπ±π8,nZ

x=2nπ+π4 or, x=2nπ,nZ

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