Prove that $\sin$ and $\cos$ are the only two functions which satisfy $f'(x) + \int_0^x f(t)\ dt = 0$for all real numbers $x$. Prove that $\sin$ and $\cos$ are the only two functions which satisfy $$f'(x) + \int_0^x f(t)\ dt = 0$$
for all real numbers $x$.
we do not know that $f(x$) is twice differentiable
tried: $$F(x) = \int_0^x f(t)\ dt$$
$$F'(x) = f(x)$$
$$F''(x)= f'(x)$$
so, writing the original formula in terms of $F(x)$ we have
$$F''(x)+F(x)=0$$
$$F(x) = a \cos x + b  \sin x$$ where $a, b$ = $F'(0)$and $F(0)$ respectively
$$= f(0) \cos x + F (0) \sin x$$
am i right up to here? where do i go from here?

All of the previous answers start by taking a derivative (fully justified by copper.hat and others in comments), and then solving a differential equation.
But there is another issue here:  Taking the derivative of such an equation can introduce extraneous solutions.
Here's a stupid example:  The equation $y = 0$ has a unique solution, $f(x) = 0$.  On the other hand, we can differentiate this, getting the differential equation $y' = 0$.  This, now, has solution $f(x) = C$ for any constant.  Of course, this only solves the original equation when $C = 0$.
A similar thing occurs on this problem.  Yes, the general solution to the differential equation everyone has obtained is $f(x) = \alpha \sin x + \beta \cos x$ for some complex coefficients $\alpha$ and $\beta$.
But checking the original condition, with $x=0$ plugged in, we get $\alpha\cos(0) -\beta\sin(0) = 0$, which forces $\alpha = 0$.
So, the final answer is that the solutions all have the form $f(x) = \beta \cos x$.  One can easily check that all of these do in fact solve the problem.
