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

Number of integral solution(s) of the inequality 2sin2x5sinx+2>0 in x[0,2π] is-

A
3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
5
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
6
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 3
Given
2sin2x5sinx+2>0
(sinx2)(2sinx1)>0
2sinx1<0
sinx<12
1sinx<12
Hence, The integral solution is 1,0
x=(0,3π2,2π)
Hence, The number of integral solution = 3

1237414_1194740_ans_3adebe15aa17495f8197850ad873c9e7.JPG

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