CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The number of values of α in [0,2π] for which 2sin3α7sin2α+7sinα=2, is :

A
6
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 3

2sin3α7sin2α+7sinα=2
2sin3α7sin2α+7sinα2=0
2(sinα1)(sin2α+sinα+1)7sinα(sinα1)=0
sinα=1 or sinα=12sinα=2
Value '2' is not possible
So, total 3 solutions α=π6,5π6,π2.


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