It doesn't make sense (to me at least!) to talk about "amplitude" or "period" of 2cos3x=1 even though I suppose I know what you mean (see 3rd post by FinalFantasy).
By some of the reasoning given above, 2cos3x=1 would have an "amplitude" of 2; yet cos3x=1/2, which is the same, would apparently have an "amplitude" of 1. Contradiction.
The solutions to 2cos3x=1 are points on a number line for the variable x. The points are the x values of where the graphs of, say, y=2cos3x and y=1 coincide (or where y=cos3x and y=1/2 coincide, or where y=2cos3x-1 and y=0 coincide, etc. All of these methods will yield the same solutions).
To systematically find the solutions of x without graphing:
2cos3x = 1
is cos3x = 1/2
Now recalling "All stations to Central" mnemonic from kindy, we carefully note that the angle 3x lies in the 1st or 4th quadrant.
3x = +- arccos(1/2) + 2n(pi) , integer n
x = +- (1/3)arccos(1/2) + (2/3)n(pi)
Correct me if I'm wrong - bit rusty coz of the summer hol.
By the way, it might be useful to remember this:
---------------------------------------
For f(x) = M+Acos(Bx+C) = M+Acos[B(x+C/B)]
the graph of y=f(x) is like the cos graph you know, except that:
Amplitude = |A|
Period = |2pi/B|
The graph is shifted up by M (or down if M is negative)
The graph if shifted left by C/B (or right if C/B is negative)
If A is negative then remeber that the shape of the graph is like upside-down cos.
----------------------------------------
phew