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

Solve the following equations.
sin2x+sin22x+sin23x+sin24x=2.

Open in App
Solution

sin2x+sin22x+sin23x+sin24x=2
Multiply 2 on both sides of the equation and use the identity cos2x=12sin2x
=>2sin2x+2sin22x+2sin23x+2sin24x=4=>1cos2x+1cos4x+1cos6x+1cos8x=4=>4(cos2x+cos4x+cos6x+cos8x)=4=>cos2x+cos4x+cos6x+cos8x=0=>cos2x+cos8x+cos4x+cos6x=0=>2cos5xcos3x+2cos5xcosx=0=>2cos5x(cos3x+cosx)=0=>4cos5x×cos2x×cosx=0cos5x=0cos2x=0cosx=0=>5x=(2n+1)π2=>2x=(2m+1)π2=>x=(2k+1)π2=>x=(2n+1)π10=>x=(2m+1)π4
n,m,k belongs to integers.

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