Express 4 sin 2x – 3cos 2x in the form of Rcos(2x+a), α>=0^0 .?

4 sin(2x) - 3 cos(2x)
= 5* [(4/5)sin(2x) - (3/5)cos(2x)]
= -5 * [(3/5)cos(2x) - (4/5)sin(2x)].
Note that 3/5 = cos(53.13 deg) = sin(53.13) deg,
so now you have:
-5 * [cos(2x)cos(53.13 deg) - sin(2x)sin(53.13)deg]
= -5 * cos(2x + 53.13 deg).
So the given equation says:
-5 *cos(2x + 53.13 deg) = 4, or
cos(2x + 53.13 deg) = -4/5, or
2x + 53.13 deg = 90 deg + 53.13 deg, or
x = 45 deg or pi/4 radians.

But that might've been obviously by "inspection" of the equation 4 sin(2x) - 3 cos(2x) = 4.

3cos 2x

Are you come from IVE?

a) 4sin(2x)-3cos(2x)=
5[4sin(2x)/5-3cos(2x)/5]=
-5[cos(a)cos(2x)-sin(a)sin(2x)]=
-5cos(2x+a)=>
R=-5
a=cos^-1(3/5)=0.93 rad. approximately.

b) -5cos(2x+0.93)=4=>
cos(2x+0.93)=-4/5=>
2x+0.93=2.5=>
x=0.785 rad.=45* approximately.