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Write #sin x +sin 3x# as a product. Hence,find all angles #x# in the interval#[0,2pi]# that satisfy #sin x +sin 3x - cos x =0#?

Answer 1

Caroline Adams

#x in{pi/12," "(5pi)/12," "pi/2," "(3pi)/2}#

First you should be aware that: #sinA+sinB-=2sin((A+B)/2)cos((A-B)/2)#

So using the above result, #sinx+sin3x=2sin((x+3x)/2)cos((x-3x)/2)#

#=2sin(2x)cos(-x)=color(blue)(2sin(2x)cos(x))#

Hence,

#sinx+sin3x-cosx=0#

Becomes,

#2sin(2x)cos(x)-cos(x)=0#

Factor out #cos(x)#

#=>cos(x)[2sin(2x)-1]=0#

Case 1 : #cos(x)=0#

Thus , #x=+-pi/2+2npi" "# , #ninZZ#

Now we substitute all the values of #n# that will give us answers in in the interval #[0,2pi]#

#n=0" " =>x=color(orange)(pi/2)# #n=1" " =>x=-pi/2+2pi=color(orange)((3pi)/2)#

Case 2: #2sin(2x)-1=0#

#=>sin(2x)=1/2#

#=>2x=pi/6(-1)^n+npi" "# , #ninZZ# #=>x=pi/12(-1)^n+npi/2#

Again, now we substitute all the values of #n# that give us the range we want,

#n=0" " =>x=color(orange)(pi/12)#

#n=1" " =>x=-pi/12+pi/2=color(orange)((5pi)/12)#

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