Equation $3x^4 + 2x^3 + 9x^2 + 4x + 6 = 0$ Solve the equation 
$$3x^4 + 2x^3 + 9x^2 + 4x + 6 = 0$$
Having a complex root of modulus $1$.
To get the solution, I tried to take a complex root $\sqrt{\frac{1}{2}} + i \sqrt{\frac{1}{2}}$ but couldn't get the solution right. Please help me.

Let the root be  $\cos y+i\sin y,$
Using Complex conjugate root theorem,  $\cos y-i\sin y$ must be another root
So, if the four roots are $\cos y\pm i\sin y,u, v$
using Vieta's formula, $(\cos y+i\sin y)(\cos y-i\sin y)u\cdot v=\dfrac63\implies v=\dfrac2u$
So we have $$3[x-(\cos y+i\sin y)][x-(\cos y-i\sin y)](x-u)\left(x-\dfrac2u\right)=3x^4 + 2x^3 + 9x^2 + 4x + 6$$
$$\iff3[(x^2-2x\cos y+1)]\left[x^2-\left(u+\frac2u\right)+2\right]=3x^4 + 2x^3 + 9x^2 + 4x + 6$$
$$\iff3\left[x^4-x^3\left(2\cos y+u+\frac2u\right)+x^2\left[1+2+2\cos y\left(u+\frac2u\right)\right]+\cdots\right]=3x^4 + 2x^3 + 9x^2 + 4x + 6$$
Equating the coefficients of $x^3,x^2$ 
$$2\cos y+u+\frac2u=-\frac23\ \ \ \ \ (1)$$ and $$1+2+2\cos y\left(u+\frac2u\right)=\frac93\iff2\cos y\left(u+\frac2u\right)=0$$
If $\cos y=0\implies\sin y=\pm1\implies \cos y\pm\sin y=\pm i$ which does not satisfy the given equation
So, $u+\dfrac2u$ must be $0$
and from $(1),2\cos y=-\dfrac23\iff\cos y=-\dfrac13\implies\sin y=\pm\dfrac{2\sqrt2}3$
Observation:   $x^2+2$ is a factor of $$3x^4 + 2x^3 + 9x^2 + 4x + 6$$
