Question

Solve for `xa n dy :sqrt(3)sinx+cosx=8y-y^2-18 ,w h e r e0lt=xlt=4pi, y in R`

Answer 1

R.H.S. `8y-y^(2)-18=-[y^(2)-8y]-18`
`=-[(y-4)^(2)-16]-18`
`=-2-(y-4)^(2)`
`:. R.H.S. le -2`
whereas `L.H.S. ge -2`
`:.` Equality is possible only when `L.H.S.=R.H.S.=-2`
Now, for `L.H.S., 2 cos (x-pi//3)=-2`
`:. cos (x-pi//3)=-1`
`:. x-pi//3=pi or 3pi`
`:. x=4pi//3, 10 pi//3`
For `R.H.S.=-2, y=4`.