wiz-icon
MyQuestionIcon
MyQuestionIcon
2
You visited us 2 times! Enjoying our articles? Unlock Full Access!
Question

If 0x<2π , then the number of real values of x, which satisfy the equation cosx+cos2x+cos3x+cos4x=0, is

A
3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
5
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
7
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
9
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 7
Given equation iscosx+cos2x+cos3x+cos4x=0(cosx+cos3x)+(cos2x+cos4x)=02cos2xcosx+2cos3xcosx=02cosx(cos2x+cos3x)=02cosx(2cos5x2cosx2)=0cosxcos5x2cosx2=0cosx=0or cos5x2=0or cosx2=0Now, cosx=0x=π2, 3π2 [0x<2π]cos5x2=05x2=π2, 3π2, 5π2, 7π2, 9π2, 11π2,....x=π5, 3π5, π, 7π5, 9π5 [0x<2π]and cosx2=0x2=π2, 3π2, 5π2,......x=π [0x<2π]Hence, x=π2, 3π2, π, π5, 3π5, 7π5, 9π5

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