I know that for the delta Dirac function the following property holds true: $δ(at)=\frac1{|a|}δ(t)$. Can i generalize that for any function:$f(at)=\frac1{|a|}f(t)$ ?
The equation i want to test it against is: $sin\big(2x+\fracπ4\big) = \frac12$. I solve below this equation using the above property and without using it.
solution without using the property: $f(at)=\frac{1}{|a|}f(t)$ $$ \begin{align} sin\big(2x+\fracπ4\big) &= \frac12 \,\,\,\,\,\,\,\,\,\,\,\,\,\, <=>\\ sin\big(2x+\fracπ4\big) &= sin\Big(\fracπ6\Big) <=> \\ \Big\{_{2x+\fracπ4 =\, 2kπ+π-\fracπ6}^{2x+\fracπ4 =\,2kπ+\fracπ6}=> \Big\{_{x=kπ+\frac{7π}{24}}^{x = kπ-\frac{π}{24}}\,\,\,, \small{k=integer} \end{align} $$
solution using the property: $f(at)=\frac{1}{|a|}f(t)$
$$ \begin{align} sin\big(2x+\fracπ4\big) &= \frac12 \\ sin\big( \, 2(x+\fracπ8) \,\big) &= \frac12 \\ \frac12sin\big(x+\fracπ8\big) &= \frac12 \,\,(just\,used\,the\,property)\\ sin\big(x+\fracπ8\big) &= 1 \\ sin\big(x+\fracπ8\big) &= sin(\fracπ2) <=>\\ x+\fracπ8 &= 2kπ+\fracπ2 \,\,\,\,\,\,\,\,\,=>\,\,\,x = 2kπ+\frac{3π}{8}\\ or\\ x+\fracπ8 &= 2kπ+π-\fracπ2 =>\,x = 2kπ+\frac{3π}{8}\\ \end{align} $$
If the property $f(at)=\frac1{|a|}f(t)$ holds true, then why it does not work in the calculations(when i use that property), what kind of knowledge am i missing? What am i doing wrong? Can you propose a solution(if any for the second solution where i am using the property)?
In general the property $f(at)=\frac1{|a|}f(t)$ does not hold true !
Counterexample: $f(t)=t^2$ or $f(t)=e^t$, etc ....