I'm assuming we're talking about radians here.
For those of you who haven't seen the notation, lg(x) = log2(x).
Any solution must lie between 0 (before which lg(x) is undefined), and 2 (after which lg(x) is greater than 1). sin(x) is nonnegative in this interval, so we can further eliminate solutions less than 1 (where lg(x) is negative).
The derivatives of these functions are cos(x) and 1/(x ln 2). So the slope of lg(x) at x=1 is ~1.442695. The slope is strictly decreasing from there. At 1.0, sin(x) is ~0.84147. If lg(x)'s slope never got lower, it wouldn't "catch up" to sin(x) before that function started its downswing at pi/2. To me, that means that there is only one solution, and it is between pi/2 and 2.
A quick way to find the solution is Newton's method, which is a series defined by
x(n+1) = x(n) - f(x)/f'(x).
In this case, f(x) = sin(x) - lg(x) and f'(x) = cos(x) - 1/(x ln 2).
From my initial estimates, my first estimate would be x(0) = 1.8. Plugging and chugging away:
x(1) = 1.922339
x(2) = 1.918667
x(3) = 1.918664
...
To recap, there's a single solution for x, and it lies at x = ~1.918664.
Solution
