Solve the equation for A (in degrees):

$2{\cos A}^2+3\cos A-2=0,0<A<90°$

The correct option is 60°.

We are given the trigonometric equation:

$2{\cos A}^2+3\cos A-2=0$

where $0<A<90°$.

Let $x=\cos A$. Substituting $x$ into the equation transforms it into a standard quadratic equation in terms of $x$:

$2x^2+3x-2=0$

We can solve this quadratic equation by splitting the middle term:

$2x^2+4x-x-2=0$

$2x(x+2)-1(x+2)=0$

$(2x-1)(\mathrm{x}+2)=0$

Setting each factor to zero gives two possible values for $x$:

$2x-1=0\Rightarrow x=\frac{1}{2}$

$x+2=0\Rightarrow x=-2$

Now, substitute back $x=\cos A$:

1. Case 1: $\cos A=-2$

Since the value of cosine for any real angle must lie within the range $[-1,1]$, $\cos A=-2$ is impossible and has no real solution.

2. Case 2: $\cos A=\frac{1}{2}$

Given that $0<A<90°$, the unique angle in the first quadrant satisfying $\cos A=\frac{1}{2}$ is:

$A=60°$

Thus, the value of A is 60°.