Let $z=\cos\theta + i\sin\theta$, then the give equation is equivalent to \begin{align*} &\frac{1}{2}\times\Big((z+\frac{1}{z})+(z^2+\frac{1}{z^2})+(z^3+\frac{1}{z^3})\Big)\\ =&-\frac{i}{2}\times\Big((z-\frac{1}{z})+(z^2-\frac{1}{z^2})+(z^3-\frac{1}{z^3})\Big) \end{align*} or \begin{align*} (z^6+z^5+z^4)(1+i) + (z^2+z+1)(1-i) &=0\\ (z^2+z+1)(z^4(1+i) + (1-i))&=0\\ (z^2+z+1)(z^4-i)&=0 \end{align*} Therefore \begin{align*} z^2 + z + 1 = 0 &\implies \theta = \boxed{\Big((2k+1)\pm\frac{1}{3}\Big)\pi}\\ z^4 - i = 0 &\implies \theta = \boxed{\Big(\frac{k}{2}+\frac{1}{8}\Big)\pi} \end{align*} where $k$ is an integer.