Note that if [tex]c=0[/tex], we have equality, but this goes against our assumption that [tex]c>0[/tex]. Note that [tex](\tanh x)'=\sech^2x>0[/tex] for all [tex]x[/tex], which means [tex]\tanh(cx)[/tex] is a monotonically increasing function, and is bounded between -1 and 1. On the other hand, [tex]-x[/tex] is a monotonically decreasing function that is unbounded. From this you can gather that the two functions never intersect for [tex]x>0[/tex] (since [tex]-x[/tex] is always negative while [tex]\tanh x[/tex] is always positive), which means [tex]c=0[/tex] is the only solution to the equation above. However, this solution is actually extraneous, since the original equation contains a factor of [tex]\dfrac1c[/tex]. So, in fact, the equation above has no solution for [tex]c[/tex].
That leaves us with [tex]a=3[/tex], [tex]b=\dfrac32[/tex], and [tex]c=2[/tex].