the equation 3sinx+cos^2x=2 is solved belowPlease help! Thanks!

we have that
3sin x+cos²x=2
3sin x+(1-sin²x)=2
-sin²x +3sin x=1
sin²x -3 sin x+1=0---------> up to here the procedure is correct
let
A=sin x
then
A²-3A+1=0
Group terms that contain the same variable, and move the constant to the opposite side of the equation(A²-3A)=-1
Complete the square. Remember to balance the equation by adding the same constants to each side.(A²-3A+2.25)=-1+2.25Rewrite as perfect squares(A-1.5)²=1.25
(+/-)(A-1.5)=1.12first solution(A-1.5)=1.12--------> A=1.12+1.5-----> A=2.62-----> sin x=2.62second solution
-(A-1.5)=1.12----> -A+1.5=1.12----> A=1.5-1.12----> A=0.38---> sin x=0.38the answer isthe equation was factored incorrectly