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

General solution of the equation 2sin2x+3cot2x4sinx6cotx+5=0 is

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

The correct option is D x=ϕ
2sin2x+3cot2x4sinx6cotx+5=0
2sin2x4sinx+2+3cot2x6cotx+3=0
2(sinx1)2+3(cotx1)2=0
sinx1=0 and cotx1=0
sinx=1 and cotx=1
But when sinx=1cotx=0 as cotx=cosxsinx
So, x=ϕ

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon