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Hi,

First, sin(4pi/3 -x) = sin(pi + pi/3 -x) = -sin(pi/3-x) = sin(x-pi/3) since sin(pi + a) = -sin(a) = sin(-a).

Now sin(x-pi/3) = sinxcos(pi/3) -cosxsin(pi/3) using formula for sin(A-B).

This=(1/2)sinx-(sqrt(3)/2)cos(x).

Using this in the given LHS, sqrt(3)sin(4pi/3 -x) + 2cosx, we get (sqrt(3)/2) sinx - (3/2)cosx + 2cosx = (sqrt(3)/2) sinx + (1/2) cosx

This is of the form sinx cosA + cosx sinA = sin(x+A) where A=30 deg +any multiple of 360 deg= pi/6 + 2pi*n where n is any integer.

So we get sin(x+pi/6+2npi) =1 (the RHS of the original question being 1)

The sine of what angle is 1? The answer is 90 deg, 360 + 90 deg, 720 + 90 deg etc, i.e. pi/2 + 2mpi where m is again any integer.

So we have that x + pi/6 + 2npi = pi/2+2mpi, or x = pi/3 +2*(m-n)*pi. However, since m and n are arbitary integers, so is their difference, which we'll call r.

So the general solution is x = pi/3 + 2rpi, r is any integer. In particular, x= pi/3 satisfies the given equation.