Question

If the equation x${}^2$ + 12 + 3 sin(a + bx) + 6x = 0 has atleast one real solution where $a,b\in \left[0,2\pi \right]$, then value of cos$\theta$ where $\theta$ is least positive value of a + bx is

• A.
• B. 2
• C.

Answer 1

${\left(x+3\right)}^2+3+3\sin \left(a+bx\right)=0$
x = - 3, sin(a + bx) = - 1
$\Rightarrow \sin \left(a-3b\right)=-1$
$a-3b=\left(4n-1\right)\frac{\pi }{2},n\in 1$
n = 1
a - 3b = 3$\pi$/2
cos(a - 3b) = 0