Question 1027793
{{{4cos^2(x) + 3sin(2x) = 4((1+cos2x)/2)+3sin2x }}}

= {{{2 + 2cos2x +3sin2x = 2+sqrt(13)((2/sqrt(13))cos2x + (3/sqrt(13))sin2x)}}}

= {{{2+sqrt(13)(cosy*cos2x + siny*sin2x) }}}

={{{ 2+sqrt(13)(cos(2x-y))}}}, where

{{{cosy = 2/sqrt(13)}}}, {{{siny = 3/sqrt(13)}}}, and {{{y = tan^-1(3/2)}}}.

Therefore,

A = 2, {{{R = sqrt(13)}}}, and {{{y = tan^-1(3/2)}}}.