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I know that my reasoning is incorrect, I just don't know where I went wrong. I did discuss this with my Maths teacher, and even she could not find what I did wrong.

Let us begin by assuming a function, $f(x)$ that is continuous and has an antiderivative in the interval $[0, 2\pi]$. Let $A$ be the area under the curve for $f(x)$ in the interval $[0, 2\pi]$

$A = \displaystyle \int_{0}^{2\pi}{f(x)\space\mathrm{d}x}$

Now there must exist a function, $g(x)$ such that:

$f(x) = g(x)\cdot \cos(x)$

Substituting the value of $f(x)$:

$A = \displaystyle\int_{0}^{2\pi}{g(x)\cdot \cos(x)\space\mathrm{d}x}$

Using t substitution:
Let $t = \sin(x)$
Then: $\mathrm{d}t = \cos(x)\space\mathrm{d}x$
And: $x = \arcsin(t)$

Changing the limits:
$t = \sin(x)$
$0$ becomes $\sin(0) = 0$
$2\pi$ becomes $\sin(2\pi) = 0$

Substituting in the definite integral:

$A = \displaystyle \int_{0}^{0}{g(\arcsin(t))\space\mathrm{d}t}$

But Definite Integral where the lower and upper bounds are the same is $0$.
So:

$A = 0$, which is not possible.

Thanks for the help.