Find the small positive numberxsuch that

cos(x^2+2x+9)=0

Round your answer to 4 decimal places.

So far I've tried this equation: x^2+2x+(9-(2n+1)pi/2)=0

put it into quadratic and for that i got x=-4.8688x10^-4 which wasn't correct

Also tried: x^2+2x+(9-pi/2)=0

got 1.9033 which also wasn't correct

T_T;; please help I only have 1 more chance or I won't get full credit. I am totally lost, I have NO IDEA what to do...but I did try.