CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
5
You visited us 5 times! Enjoying our articles? Unlock Full Access!
Question

The set of values of λ such that the equation cosθ+cos2θ+λ=0 admits of a solution for θ is

A
[0,+)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
[0,98]
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
[3,0]
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
None of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B [0,98]

cosθ+cos2θ+λ=0
cosθ+2cos2θ1+λ=0
2cos2θ+cosθ+(λ1)=0
cosθ=1±98λ4
We know,
98λ0
λ98
1cosθ1
41±98λ4
998λ
λ0
so,
[0,98]


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