Trigonometric roots in Complex Plane!

Since $x=\cos{θ}+i\sin{θ}$ and $x=\sin{θ}+i\cos{θ}$ are roots, then $x=\cos{θ}-i\sin{θ}$ and $x=\sin{θ}-i\cos{θ}$ are also roots for there to be real coefficients for $P(x)$ . The $y$ -intercept of $P(x)$ is the product of the $4$ roots which is $(\cos^2{θ}+\sin^2{θ})^2$ which is $1$ . When the roots are graphed onto the complex plane, a Trapezoid can be formed with bases $2\cos{θ}$ and $2\sin{θ}$ and height $\sin{θ}-\cos{θ}$ since $\sin{θ}>\cos{θ}$ . The area of the Trapezoid comes out to be $\sin^2{θ}-\cos^2{θ}$ which is half the $y$ -intercept of $P(x)$ or $0.5$ . Thus, $1-2\cos^2{θ}=0.5$ . Solving this yields $\cos{θ}=0.5$ and $\sin{θ}=0.5\sqrt{3}$ as $\sin{θ}>\cos{θ}>0$ . The coefficient of the $x^3$ term is the negative of the sum of roots which is $-2\cos{θ}-2\sin{θ}$ or $-1-\sqrt{3}$ . Thus, $a=-1$ and $b=3$ so $2(-1)-3(3)+18=\boxed{7}$ which is the final answer.