Show that if $a\sin x + b\cos x + ce^x$ is the zero function for constants $a, b, c$ then $a = b = c = 0$ 
Suppose that $a\sin x + b\cos x + ce^x$ is the zero function. Prove that $a=b=c=0$.

Does the zero function simply mean that $a\sin x + b\cos x + ce^x = 0$? I am going under the assumption that it is.

Geometrically speaking, I think it is pretty clear that $a=b=c=0$ must be true if $a\sin x + b\cos x + ce^x = 0$ as there are no x-values that make the equation true otherwise.
Is there a purely algebraic way of saying this that does not utilize showing a graph?
 A: Let $x=0$.
Thus, $$b+c=0$$
Let $x=\frac{\pi}{2}.$
Thus, $$a+c\cdot e^{\frac{\pi}{2}}=0.$$
Now, substitute $x=-\frac{\pi}{2}$ and solve this system.
We obtain: $$-a+c\cdot e^{-\frac{\pi}{2}}=0.$$
Thus,
$$c\left(e^{\frac{\pi}{2}}+e^{\frac{\pi}{2}}\right)=0$$ or
$$c=0,$$ which gives $a=0$ and $b=0$.
A: The Wronskian (determinant) addresses exactly this question, and applies just as well to an arbitary number of ($(n - 1)$-times differentiable) functions: The functions $f_1, \ldots, f_n$ are linearly independent if and only if the determinant $$W[f_1, \ldots, f_n]$$ of the matrix
$$\pmatrix{f_1 & \cdots & f_n \\ f_1' & \cdots & f_n' \\ \vdots & & \vdots \\ f_1^{(n - 1)} & \cdots & f_n^{(n - 1)}}$$ is identically zero.
In our case, $f_1 = \sin, f_2 = \cos, f_3 = \exp$, and computing directly gives
$$W[\sin, \cos, \exp](x) = -2 e^x ,$$
which is not (identically) zero, so the three functions are linearly independent, that is, the only solution $(a, b, c)$ to $$a \sin x + b \cos x + c e^x = 0$$ the zero solution.
A: Given
$a\sin x + b\cos x + ce^x = 0, \tag 1$
we multiply through by $e^{-x}$, which yields
$ae^{-x}\sin x + be^{-x}\cos x + c = 0; \tag 2$
letting $x \to \infty$, we find that
$c = 0; \tag 3$
we are left with
$a\sin x + b\cos x = 0; \tag 4$
now set
$x = \dfrac{\pi}{2}, \tag 5$
and obtain
$a = a\sin \dfrac{\pi}{2} + b\cos \dfrac{\pi}{2} = 0; \tag 6$
thus
$a = 0; \tag 7$
in a similar manner, choosing
$x = 0 \tag 8$
we find
$b = a \sin 0 + b \cos 0 = 0 \tag 9$
as well.  
We have in fact established that $\sin x$, $\cos x$, and $e^x$ are linearly independent.
A: If you expand sin, cos and exp into series, and then add them and get one series for the sum, then function is 0 if and only if coefficient on every x^n will be 0, and thus a=b=c=0 will be the only solution for that.
A: Assume $c \neq 0$. Then $x \longmapsto e^x$ is a linear combination of $\sin$ and $\cos$, thus is periodic. It is impossible.
Thus $c=0$, and $a\cos +b\sin=0$. Then evaluate at $0$ and $\pi/2$. 
