Solving a Fourier sine transform equation

Suppose we have the following Fourier sine transform equation $$\int_0^\infty f(x;p)\sin(\lambda x)dx \equiv 0,$$ where $f(x;p)$ has some parameters $p\in\mathbb{R}$ we can choose freely. Does this imply $f(x;p)\equiv 0$? N.B. the symbol $\equiv$ is used to mean identical.

My reasoning goes as follows: We have $F_s(\lambda;p)\equiv 0$ by definition of the integral equation. Applying the inverse Fourier sine transform gives

$$f(x;p) = \sqrt{\frac{2}{\pi}}\int_0^\infty F_s(\lambda;p)\sin(x \lambda)d\lambda.$$ But this implies $f(x;p)\equiv 0$ since $F_s(\lambda;p)\equiv 0$. Is this correct, or are there any special conditions that $F_s(\lambda;p)$ cannot be identically zero when applying the inverse Fourier sine transform?

Your problem reduces to the following statement:

"$\int_0^\infty g(x)\sin(\lambda x)dx=0\quad\forall \lambda\in \Bbb R.$ Does it follow that $g$ is identically zero?"

Suppose that $g$ is real-valued. We pose $g(x)=0$ for negative $x$. Then we have the following identities:

$$\int_0^\infty g(x)\sin(\lambda x)dx=\Im (\mathcal F[g](\lambda))=0.$$

$\Im$ stands for "imaginary part".

Similarly, $\Im\mathcal F[\bar g]=-\Im \mathcal F[ g]$. where $\bar g(x)=g(-x)$. Therefore we can say $$\mathcal F[g-\bar g]=2\Im \mathcal F[g]=0,$$ hence, $g=\bar g$ because Fourier transform is injective. We conclude immediately that $g$ is identically zero.

The proof for the case where $g$ is complex-valued is open (at least I can't find an evident way right now).