Question #8dc79

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Answer 1 · 2016-09-14T15:27:53

Let
#5cosx+3cos(x+pi/3)+3=y#

#=>5cosx+3cosxcos(pi/3)-3sinxsin(pi/3)+3=y#

#=>5cosx+3/2cosx-(3sqrt3)/2sinx=y-3#

#=>13/2cosx-(3sqrt3)/2sinx=y-3#

Now dividing both sides by #sqrt((13/2)^2+((3sqrt3)/2)^2)=sqrt(169/4+27/4)#
#=sqrt((169+27)/4)=7# we get

#13/14cosx-(3sqrt3)/14sinx=(y-3)/7#

Now if #13/14=cosA" then "(3sqrt3)/14=sinA#

So

#cosAcosx-sinAsinx=(y-3)/7#

#=>cos(x+A)=(y-3)/7#

Now as #-1<=cos(x+A)<=+1# we can write

#-1<=(y-3)/7<=+1#

#=>-7<=(y-3)<=+7#

#=>-7+3<=(y-3+3)<=+7+3#

#=>-4<=y<=+10#

So the given expression
lies between -4 and 10.

Proved