you might want to say
let cosØ = x, then
x^2 + x - 1 = 0
x = (-1 ± √5)/2 = .618033 or -1.618..
cosØ = .618033 or cosØ = -1.6.. , which is not possible since -1 ≤ cosx ≤ 1
and since our domain is between 0° and 180°, we only need the angle in quadrant I
so Ø = 51.83°
Solve each equation for the given domain.
cos^2 theta + cos theta - 1 = 0 for 0-180.
1 answer
1 answer