In the vector space, $V =\{f :\mathbb R \longrightarrow\mathbb R\}$, prove that the set $\{\cos 2x, \sin 6x, x^3\}$ is linearly independent. In the vector space, $V = \{f : \mathbb R \longrightarrow\mathbb R\}$, prove that the set $\{\cos{2x},\sin{6x},x^3\}$ is linearly independent.
I’ve started with $a\cos{2x} + b\sin{6x} + cx^3=0$, trying to prove that $a=b=c=0$
I know when I let $x=0$ I find $a=0$ but I'm unsure where to go from there. 
My guess is that the set is linearly dependent.
 A: You did well when you started with the equality $a\cos(2x)+b\sin(6x)+cx^3=0$. Now, taking $x=0$, you deduce right away that $a=0$. So, now you just have $b\sin(6x)+cx^3=0$. Take $x=\pi$. Then you deduce that $c=0$. And, of course, the only $b$ for which $b\sin(6x)$ is the null function is $b=0$.
A: The method in the other answers, choosing specific values of $x$ in order to deduce one at a time that $a$, $b$, and $c$ are all $0$, works just fine and has the virtue of working directly with the definition of linear independence. I'm writing this answer just to provide two alternative perspectives:

First, you can look at the three functions in order, and ask of each whether it is a linear combination of the previous functions: is it "redundant"? In other words, we're trying to check that the dimension of the vector subspace they generate increases by one. 
For the first, $\cos(2x)$, it's not identically zero so it generates a dimension-one subspace by itself. (It's not a linear combination of the zero functions before it in the list.)
For the second, $\sin(6x)$, it's not always zero when $\cos(2x)$ is zero, so it can't be a multiple of $\cos(2x)$. So $\cos(2x)$ and $\sin(6x)$ generate a two-dimensional subspace. 
Finally, is $x^3$ a linear combination of $\cos(2x)$ and $\sin(6x)$? No: one reason is that $\cos(2x)$ and $\sin(6x)$ are both periodic and repeat every $2\pi$ units, so every linear combination is too, while $x^3$ is not. Another reason is that $\cos(2x)$ and $\sin(6x)$ are both bounded, so every linear combination of them is too, but $x^3$ is not. So $\cos(2x)$, $\sin(6x)$, and $x^3$ together generate a three-dimensional subspace, and they must all be linearly independent.

A determinantal perspective is that for smooth functions $f$, $g$, and $h$, if they have some relation $af(x)+bg(x)+ch(x)=0$, then their derivatives have the same relation $af'(x)+bg'(x)+ch'(x)=0$, and so do their second derivatives $af''(x)+bg''(x)+ch''(x)=0$. That means the matrix equation
$$ \begin{pmatrix} f(x) & g(x) & h(x)\\f'(x) & g'(x) & h'(x) \\ f''(x) & g''(x) & h''(x) \end{pmatrix}\begin{pmatrix} a\\ b\\ c\end{pmatrix} = \begin{pmatrix} 0\\0\\0\end{pmatrix} $$
holds identically, so if $a$, $b$, and $c$ aren't all zero then that $3\times 3$ matrix has nontrivial kernel and must have determinant $0$ identically.
This determinant is called the Wronskian of $f$, $g$, and $h$. If you calculate it for $\cos(2x)$, $\sin(6x)$, and $x^3$ you get one, single, complicated function that you only have to check is nonzero in one place to know that they are linearly independent. This technique is especially helpful to show that you've found a basis for all the solutions to a degree-$n$ differential equation, which is known to be $n$-dimensional, so if you have $n$ linearly independent solutions then they must form a basis.
A: Consider
$$\forall x\in \Bbb{R}, a\cos{2x}+b\sin{6x}+cx^3=0$$
As you pointed out by letting $x=0$ you get $a=0$. With $x=\pi$ you get $c\pi^3=0$ i.e $c=0$ (Keep in mind $a=0$)
Now take $x=\pi/12$ you get $b=0$ (Keep in mind $a=c=0$)
A: Assume if possible they are linearly dependent. Now it is obvious that $\sin 6x$ and $\cos 2x$ are linearly independent (cause otherwise, $\sin 6x$ would have been equal to a constant times $\cos 2x$, for all $x$, which is certainly not the case). Now if {$x^3, \sin 6x, \cos x$} are linearly dependent, then for all $x, x^3$ can be represented as a linear combination of $\sin 6x$ and $\cos 2x$. But it is clear that a linear combination of $\sin 6x$ and $\cos 2x$ is again sinusoidal and can not be equal to $x^3$, thus we have arrived at a contradiction.
