Is it possible to figure out how many carbon-carbon double bonds are present using just the molecular formula?
Both you and Rapid99 have missed the word "acyclic" in the problem statement.  So you don't have to worry about rings (because they said acyclic).  [Also no triple bonds.]

You also know that it's a carboxylic acid ("COOH group", formula but not structure of carboxylic acid group), so don't worry about something funky with the oxygens.  The formula is then $\ce{C19H31-COOH}$.

If the part of the compound on left was full saturated and sans $\ce{COOH}$ it would be $\ce{C19H_{(2*19+2)}}=\ce{C19H40}$.  Remember any acyclic alkane has formula $\ce{C_nH_{2n+2}}$.

Accounting for the $\ce{COOH}$ group location means the saturated alkane would then be $\ce{C19H39-COOH}$ (take away one $\ce{H}$ "spot" to attach the COOH group).

So we are really 8 hydrogens short: 39-31 (from above) = 8.  Each double bond takes away two hydrogens.  So if there are 8 missing hydrogens, there must be 4 double bonds.  X=8/2.