Mathematics - Trigonometric Equation Question with Solution | TestHub

The given equation is $x^2+6x+12+3\sin (a+bx)=0$.

Rewrite it as $(x+3{)}^2+3+3\sin (a+bx)=0$.

For a real solution, $(x+3{)}^2\ge 0$ and $3+3\sin (a+bx)\ge 0$.

Thus, $(x+3{)}^2=0$ and $3+3\sin (a+bx)=0$.

This implies $x=-3$ and $\sin (a+bx)=-1$.

So, $\sin (a-3b)=-1$. The least positive value of $a+bx$ (which is $\theta$) for $\sin (\theta )=-1$ is $\theta =\frac{3\pi }{2}$.

Therefore, $\cos \theta =\cos (\frac{3\pi }{2})=0$.

$\boxed{C}$