Hi, I'm stuck at this question:
List all points on the graph of y=tanx on the interval [-3π/2,2π] that have a y-coordinate of (1/radical3).
This is my answer, but I keep getting it wrong. Someone, please explain how to go about solving this. I have more of similar questions to solve. So, if anyone can please show work + answer so that I can use this example as a reference for others. Thank you.
-My Answers: (5π/6,1/radical3),(11π/6,1/radical3), (17π/6,1/radical3)
You did not read the domain carefully, nor did your points meet the requirements.
All of your points are in QII and QIV, where tan(x) = -1/√3
The domain is [-3π/2,2π], and all of your points must lie in QI or QIII, where tan(x) > 0
That would mean that the x-coordinates are
-5π/6, π/6, 7π/6
To find all the points on the graph of y = tan(x) on the interval [-3π/2, 2π] that have a y-coordinate of 1/√3, you need to remember the definition of the tangent function and its periodicity.
The tangent function is defined as tan(x) = sin(x)/cos(x). So, for a given x, we can find the y-coordinate by calculating sin(x)/cos(x).
In this case, we are looking for points where the y-coordinate is 1/√3. Therefore, we need to solve the equation sin(x)/cos(x) = 1/√3.
Let's simplify this equation by multiplying both sides by √3 to eliminate the fraction. We get:
(√3*sin(x))/cos(x) = 1
Now, we can simplify further using trigonometric identities. Recall that √3*sin(x) = sin(π/3 + x), and cos(x) = cos(x). Substituting these identities into the equation gives:
(sin(π/3 + x))/cos(x) = 1
To solve this equation, we can use the concept of reference angles. Since we're looking for solutions on the interval [-3π/2, 2π], we can restrict our attention to the interval [0, 2π]. Finding an angle that satisfies the equation within this interval will give us a range of solutions.
Now, we know that the reference angle for the tangent function is π/6. This means that any angle that differs from π/6 by a multiple of π will have the same tangent value. So, we'll solve the equation for angles that are π/6 more than a multiple of π.
π/3 + x = π/6 + πn, where n is an integer.
Simplifying this equation gives:
x = π/6 - π/3 + πn = -π/6 + πn
Now, we can use the interval [-3π/2, 2π] to find the values of n that give us angles within this range.
For n = -3, x = -π/6 - 3π = -3π/2
For n = -2, x = -π/6 - 2π = -π/6
For n = -1, x = -π/6 - π = -7π/6
For n = 0, x = -π/6
For n = 1, x = -π/6 + π = 5π/6
For n = 2, x = -π/6 + 2π = 11π/6
For n = 3, x = -π/6 + 3π = 17π/6
So, the points on the graph of y = tan(x) within the given interval that have a y-coordinate of 1/√3 are:
(5π/6, 1/√3)
(11π/6, 1/√3)
(17π/6, 1/√3)
Therefore, your answers are correct!