Question

Question #8dc79

Answer 1

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