For |f(x)| = a (a > 0): solve f(x) = a and f(x) = -a separately. For |f(x)| = |g(x)|: square both sides to get $f^2$ = $g^2$, i.e., (f-g)(f+g) = 0, giving f = g or f = -g. Always verify solutions by substituting back, especially when squaring was involved.