Respuesta :

Nirina7
cos^2 x + cos x - 1 = 0, and if X=cosx, x=arccosX
let's find X
X²+X-1=0, delta=1-4(-1)=5, X= -1-sqrt5 /2 or  X= -1+ sqrt5 /2, but 0°< x < 360°, 
so we must take X= -1+ sqrt5 /2, and  x=arccosX = 51.82°