Solve each equation for the given domain.
cos^2 theta + cos theta - 1 = 0 for 0-180.
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°
To solve the equation cos^2 theta + cos theta - 1 = 0 for the given domain 0-180, we can use the quadratic formula.
The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x can be found using the formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, we have cos^2 theta + cos theta - 1 = 0, meaning a = 1, b = 1, and c = -1.
Substituting these values into the quadratic formula, we have:
theta = (-1 ± √(1 - 4(1)(-1))) / (2(1))
Simplifying further:
theta = (-1 ± √(1 + 4)) / 2
theta = (-1 ± √5) / 2
Now we have two possible solutions:
1. theta = (-1 + √5) / 2
2. theta = (-1 - √5) / 2
To find the solutions within the given domain 0-180, we need to check if the solutions fall within this range.
1. theta = (-1 + √5) / 2 ≈ 0.618
Since this solution is less than 0, it does not fall within the given domain.
2. theta = (-1 - √5) / 2 ≈ -1.618
This solution is less than 0 and also does not fall within the given domain.
Therefore, there are no solutions to the equation cos^2 theta + cos theta - 1 = 0 within the domain 0-180.