Solve and discuss according to the real parameter "#a#" ? #sin(x)+a·sin(2x)+sin(3x)=0#

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Feb 9, 2018

Answer:

#a# ranges in the interval #[-2,2]# and
#x=(npi)/2# and #x=2npi+-cos^(-1)(-a/2)#

Explanation:

let us first simplify it.

As #sinx+asin2x+sin3x=0#, we can write it as

#asin2x+2sin((3x+x)/2)cos((3x-x)/2)=0#

or #asin2x+2sin2xcosx=0#

or #sin2x(a+2cosx)=0#

If #sin2x=0# then #2x=npi# and #x=(npi)/2#

if #a+2cosx=0#, then #cosx=-a/2# and as #cosx# ranges in the interval #[-1,1]#, #a# ranges between #[-2,2]# and

#x=2npi+-cos^(-1)(-a/2)#

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