Question #6d317

1 Answer
Sep 7, 2016

#x = n pi/3# (#n in ZZ#)
#color(white)("XXX")#or
#x = -pi/3+n * (2pi)/3# (#n in ZZ#)

Explanation:

Here are some formulae that you will need:
#color(white)("XXX")sin(A+B)=sin(A)cos(B)+sin(B)cos(A)#
#color(white)("XXXXXX")rarr sin(2x)=2sin(x)cos(x)#

#color(white)("XXX")cos(A+B)=cos(A)cos(B)-sin(A)sin(B)#
#color(white)("XXXXXX")rarr cos(2x)=2 cos^2(x)-1#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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

#color(green)(sin(3x)=sin(x)cos(2x)+sin(2x)cos(x))#

#color(white)("XXXX")color(green)(=sin(x)[2cos^2(x)-1]+2sin(x)cos(x)[cos(x)])#

#color(white)("XXXX")color(green)(=2sin(x)cos^2(x)-sin(x)+2sin(x)cos^2(x)#

#color(white)("XXXX")color(green)(=sin(x)[4cos^2(x)-1]#

Given
#color(white)("XXXX")color(blue)(sin(2x))+color(green)(sin(3x))=color(red)(sin(x))#

#color(white)("XXXX")color(blue)(2sin(x)cos(x))+color(green)(sin(x)[4cos^2(x)-1])=color(red)(sin(x))#

#rArr#

#color(white)("XXXX")color(purple)(sin(x)) * [4cos^2(x)+2cos(x)-1]=color(purple)(sin(x)) * [1]#

Either

#{: (sin(x)=0,color(white)("XX")"or"color(white)("XX"),4cos^2(x)+2cos(x)-1=1), (rarr x in npi,, 2cos^2(x)+cos(x)-1=0), (,,cos(x)=(-1+-sqrt(1^2-4(2)(-1)))/(2(2))), (,,cos(x)=-1" or " 1/2), (,,x in {pi+n * 2pi, -pi/3+n*(2pi)/3}) :}#