CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question

Solve for x, (πxπ) the equation : 2(cosx+cos2x)+sin2x(1+2cosx)=2sinx.
How many distinct values of x satisfy the equation in the above range

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

The correct option is C 5

Given, 2(cosx+cos2x)+sin2x(1+2cosx)=2sinx,for(πxπ),2(cosx+2cos2x1)+2sinxcosx(1+2cosx)=2sinx,(cosx+2cos2x1)+sinx(cosx+2cos2x1)=0(cosx+2cos2x1)(1+sinx)=0,sinx=1,xϵ{π2}cosx+2cos2x1,(2cosx1)(cosx+1),xϵ{π,π3,π3,π}i.e,5solutions,xϵ{π,π2,π3,π3,π}


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