For real numbers $x$, let 
\[P(x)=1+\cos(x)+i\sin(x)-\cos(2x)-i\sin(2x)+\cos(3x)+i\sin(3x)\]
where $i = \sqrt{-1}$. For how many values of $x$ with $0\leq x<2\pi$ does 
\[P(x)=0?\]
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\  1 \qquad\textbf{(C)}\  2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4$
Let $a=\cos(x)+i\sin(x)$. Now $P(a)=1+a-a^2+a^3$. $P(-1)=-2$ and $P(0)=1$ so there is a real root $a_1$ between $-1$ and $0$. The other $a$'s must be complex conjugates since all coefficients of the polynomial are real. The magnitude of those complex $a$'s squared is $\frac{1}{a_1}$ which is greater than $1$. If $x$ is real number then $a$ must have magnitude of $1$, but none of the solutions for $a$ have magnitude of $1$, so the answer is $\boxed{\textbf{(A)}\ 0 }$ ~lopkiloinm