Linear Transformation Regarding A Vector Space of Continuous Functions The question reads:

Let $V$ be the subspace of $C[0,2\pi]$ spanned by the vectors $1, \sin{x}, \cos{x}$, and let $T: V \to \mathbb{R^3} $ be the evaluation transformation at the sequence of points $0, \pi, 2\pi$. Find: $T(1+\sin{x}+\cos{x}),$ ker($T$), and the range of $T$.

First off, I want to make sure I am interpreting $V$ correctly, in that it is the set of functions that are continuous in the closed interval of $0$ to $2\pi$ and that these functions are of the form $f(x) = c_1 + c_2\sin{x} + c_3\cos{x}$ for some $c_1, c_2, c_3 \in \mathbb{R}$.
Finding $T(1+\sin{x}+\cos{x})$ seemed simple enough, and my answer is the point $(2,0,2)$.
For ker($T$), because any linear combination of $\sin{x}$ will give you zero when you evaluate it at the three points, I said that $\sin{x}$ is in the kernal of $T$. So my answer is ker($T$) = {$0,\sin{x}$}.
And as for the range, because you could have $c_2, c_3 = 0$, the function can be any constant, so all the points in $\mathbb{R^3}$ could be hit. So I said the range of $T$ is $\mathbb{R^3}$.
Is my thought process correct? Any help or criticism is appreciated.
 A: Your reasoning for the kernel and range are not correct.  For the kernel, you have shown that $\sin x$ is an element of the kernel, but there may be other elements besides $\sin x$ (and $0$).  In fact, for any $c$, $c\sin x$ must also be in the kernel since the kernel is a subspace.  This still may not be the entire kernel though: you need to determine whether it is possible for $f(x) = c_1 + c_2\sin{x} + c_3\cos{x}$ to be in the kernel if $c_1$ or $c_3$ is nonzero (and if it is possible, when exactly it happens).
For the range, you have shown that for any $a\in\mathbb{R}$, there is some point in the range which has $a$ as one of its coordinates (namely $(a,a,a)$).  However, this does not imply the range is all of $\mathbb{R}^3$: for instance, how do you know $(0,1,2)$ is in the range?  In fact, the range cannot be all of $\mathbb{R}^3$: the space $V$ has dimension at most $3$ (it is spanned by three elements) and $\ker T$ has dimension at least $1$ (it contains $\sin x$), so the dimension of the image of $T$ is at most $3-1=2$.
