find, to the nearest minute, the solution set of 4 cos x + 5 = 6 sec x over the domain 0 degrees
To find the solution set of the equation 4cos(x) + 5 = 6sec(x) over the domain 0 degrees, we can use trigonometric identities and algebraic manipulations.
First, let's manipulate the equation to eliminate the sec(x) term. We know that sec(x) is the reciprocal of cos(x), so we can rewrite the equation as:
4cos(x) + 5 = 6/cos(x)
Multiply both sides by cos(x) to eliminate the fraction:
4cos^2(x) + 5cos(x) = 6
Rearrange the terms to get a quadratic equation in terms of cos(x):
4cos^2(x) + 5cos(x) - 6 = 0
To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula. Let's use the quadratic formula:
cos(x) = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 4, b = 5, and c = -6:
cos(x) = (-(5) ± √((5)^2 - 4(4)(-6))) / (2(4))
Simplifying further:
cos(x) = (-5 ± √(25 + 96)) / 8
cos(x) = (-5 ± √121) / 8
cos(x) = (-5 ± 11) / 8
Now we have two possible solutions for cos(x):
1. cos(x) = (6) / 8 = 3/4
2. cos(x) = (-16) / 8 = -2
To find the corresponding values of x in degrees, we can use the inverse cosine function. However, we need to consider the given domain of 0 degrees.
The domain of 0 degrees means we are considering angles between 0 and 360 degrees. Let's find the angles in this domain that have cos(x) = 3/4 and cos(x) = -2.
Using a calculator, we find that cos^-1(3/4) ≈ 41.41 degrees. However, this angle is outside the given domain since it is greater than 0 degrees. So we discard this solution.
For cos(x) = -2, there are no angles in the domain of 0 degrees that satisfy this equation. Therefore, there are no solutions for this case.
Hence, in the given domain of 0 degrees, the solution set of the equation 4cos(x) + 5 = 6sec(x) is empty or has no solutions.