Given 2secant^2 (theta) + 1 = 5 secant (theta), solve for (theta), to the nearest degree, in the interval 0 degrees is greater than or equal to (theta) which is less than 360 degrees
let sec(theta) = x
so we have
2secant^2 (theta) + 1 = 5 secant (theta) becomes
2x^2 + 1 = 5x
2x^2 - 5x + 1 = 0
x = (5 ± √17)/4
= 2.2808 or .21922
so sec(theta) = 2.2808
cos(theta) = 1/2.2808 = .43844 , so theta is in I or IV
theta = 63.995 or 64º or theta = 296º
or
sec(theta) = .21922
cos(theta) = 1/.21922 = 4.56 ---> not possible
theta = 64º or 296º
To solve the equation 2sec²(theta) + 1 = 5sec(theta) for (theta), we can first simplify it by substituting sec(theta) using the identity sec(theta) = 1/cos(theta).
The equation becomes:
2(1/cos²(theta)) + 1 = 5(1/cos(theta))
To eliminate the fractions, we can multiply both sides of the equation by cos²(theta):
2 + cos²(theta) = 5cos(theta)
Rearranging the equation:
cos²(theta) - 5cos(theta) + 2 = 0
This is now a quadratic equation in terms of cos(theta). To solve it, we can use the quadratic formula:
cos(theta) = (-b ± sqrt(b² - 4ac)) / 2a
In this case, a = 1, b = -5, and c = 2. Substituting these values into the quadratic formula:
cos(theta) = (-(-5) ± sqrt((-5)² - 4(1)(2))) / (2(1))
Simplifying:
cos(theta) = (5 ± sqrt(25 - 8)) / 2
cos(theta) = (5 ± sqrt(17)) / 2
To determine the values of cos(theta), we need to find the two possible solutions for theta that satisfy the given conditions (0 degrees ≤ theta < 360 degrees).
1. For cos(theta) = (5 + sqrt(17)) / 2:
Since the cosine function represents the x-coordinate of a point on the unit circle, we need to find the angle whose cosine is (5 + sqrt(17)) / 2. Using a calculator, we can find the inverse cosine (also known as arccosine) of this value:
theta = arccos((5 + sqrt(17)) / 2)
2. For cos(theta) = (5 - sqrt(17)) / 2:
Similarly, we can find the second solution using the inverse cosine:
theta = arccos((5 - sqrt(17)) / 2)
Both of these values will be in radians. To convert them to degrees, multiply by 180/π:
thetạ₁ = (180/π) * arccos((5 + sqrt(17)) / 2)
thetạ₂ = (180/π) * arccos((5 - sqrt(17)) / 2)
Therefore, the solutions for (theta) in the given interval (0 degrees ≤ theta < 360 degrees) are thetạ₁ and thetạ₂, rounded to the nearest degree.