Solve the equation for A (in degrees):
$$2 \cos^2 A + 3 \cos A - 2 = 0$$​ ($$0^\circ < A < 90^\circ.$$)

Correct option is B

Given:
The equation is:$$2 \cos^2 A + 3 \cos A - 2 = 0$$​​
​$$0^\circ < A < 90^\circ.$$​​​​
Solution:
$$2 \cos^2 A + 3 \cos A - 2 = 0$$​​
Let $$x = \cos A$$​ so the equation becomes:
​$$2x^2 + 3x - 2 = 0 \\ \ \\ 2x^2 +4x-x-2=0\\ \ \\ 2x(x+2)-1(x+2)=0 \\ \ \\ (2x-1)(x+2)=0 \\ \ \\ x = \frac12, -2$$​

x = $$= -2$$​ (This is not a valid solution since $$\cos A$$​ must lie between -1 and 1.)

So, we have $$\cos A = \frac12$$​

cos A = $$\cos 60^0$$​

A = $$60^\circ$$​

The value of A is $$60^\circ$$​

Thus, the correct option is (b) 60°