find the points on the graph of y=sec x, 0=<x<=2pi, where tangent is parallel to 3y-2x=4. use exact values.
To find the points on the graph of y = sec(x) where the tangent is parallel to the line 3y - 2x = 4, we need to find the derivative of y = sec(x) and set it equal to the slope of the line, which is 2/3.
Let's start by finding the derivative of y = sec(x):
The derivative of sec(x) can be found by using the chain rule:
dy/dx = d/dx(sec(x))
= sec(x)tan(x)
Now, we set the derivative equal to the slope of the line:
sec(x)tan(x) = 2/3
Next, we solve this equation for x. Keep in mind that we are only concerned with values of x between 0 and 2π since it is specified in the question.
sec(x)tan(x) = 2/3
Since tan(x) = sin(x)/cos(x) and sec(x) = 1/cos(x), we can rewrite the equation as:
(1/cos(x))(sin(x)/cos(x)) = 2/3
Simplifying further, we get:
(sin(x))/(cos^2(x)) = 2/3
Multiplying both sides by 3cos^2(x), we get:
3sin(x) = 2cos^2(x)
Using the identity cos^2(x) = 1 - sin^2(x), we can substitute:
3sin(x) = 2(1 - sin^2(x))
Expanding, we have:
3sin(x) = 2 - 2sin^2(x)
Rearranging and canceling common terms, we get:
2sin^2(x) + 3sin(x) - 2 = 0
Now, we can solve this quadratic equation for sin(x) by factoring or using the quadratic formula. Factoring this quadratic equation may not produce exact values, so we will use the quadratic formula:
sin(x) = (-b ± √(b^2 - 4ac))/(2a)
In this case, a = 2, b = 3, and c = -2. Plugging these values into the quadratic formula, we get:
sin(x) = (-3 ± √(3^2 - 4(2)(-2)))/(2(2))
= (-3 ± √(9 + 16))/4
= (-3 ± √25)/4
= (-3 ± 5)/4
Therefore, we have two possible solutions for sin(x): sin(x) = 2/4 = 1/2 and sin(x) = -8/4 = -2
To find the corresponding values of x, we use the inverse sine function, sin^(-1)(x):
x₁ = sin^(-1)(1/2)
x₂ = sin^(-1)(-2)
Using an inverse sine table or a calculator, we find the exact values for x₁ and x₂:
x₁ = π/6 or 30°
x₂ = 7π/6 or 210°
Now that we have the values of x, we can substitute them back into the equation y = sec(x) to find the corresponding y-values:
For x₁ = π/6:
y₁ = sec(π/6) = 2
For x₂ = 7π/6:
y₂ = sec(7π/6) = -2
Therefore, the points on the graph of y = sec(x) where the tangent is parallel to the line 3y - 2x = 4, for 0 ≤ x ≤ 2π, are (π/6, 2) and (7π/6, -2).