Use a graphing utility to solve the equation on the interval 0°< x < 360°. Express the solution (s) rounded to one decimal place.

cos^2 x + cos x - 1 = 0

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°