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