How do you solve 4sin^2x+4cosx-5=0

Answer 1

4sin^2x+4cosx-5=0

4 (1 -cos² x) +4 cos x -5 =0

4 -4cos² x +4cos x -5 =0

-4cos² x + 4 cos x -1 = 0

-4cos² x + 4 cos x = 1

-4 cos x (cos x - 1 ) =1

so

-4cos x = 1 and cos x - 1 = 1

cos x = -1/4 cos x = 2

x =104.48 not valid