For 0< x<π/2, if #sinx+sin3x=cosx+cos3x# then x = ?

For #0<x<π/2# if #sinx+sin3x=cosx+cos3x# then #x=#?

2 Answers
Aug 10, 2017

#sinx+sin3x=cosx+cos3x#

#=2sin2xcosx=2cos2xcosx#

#=2sin2xcosx-2cos2xcosx=0#

#=2cosx(sin2x-cos2x)=0#

So for #cosx = 0# there is no solution as # 0 < x < pi/2#

when #sin2x-cos2x=0#

#=>tan2x=1=tan(pi/4)#

#=>x=pi/8#

Aug 10, 2017

The solutions are #S={pi/2, pi/8}#

Explanation:

We need

#sina+sinb=2sin((a+b)/2)cos((a-b)/2)#

#cosa+cosb=2cos((a+b)/2)cos((a-b)/2)#

Therefore,

#sinx+sin3x=2sin((x+3x)/2)cos((3x-x)/2)=2sin2xcosx#

and

#cosx+cos3x=2cos((x+3x)/2)cos((3x-x)/2)=2cos2xcosx#

Therefore,

#sinx+sin3x=cosx+cos3x#

#2sin2xcosx=2cos2xcosx#

#2sin2xcosx-2cos2xcosx=0#

#2cosx(sin2x-cos2x)=0#

So,

#cosx=0#, #=>#, #x=pi/2+2kpi# and #x=3/2pi+2kpi#, #AA k in ZZ#

#sin2x-cos2x=0#, #=>#, #tan2x=1#, #=>#, #2x=pi/4+kpi#

#x=pi/8+k/2pi#, #AA k in ZZ#