List all points on the graph of y=tanx on the interval [π/2,3π] that have a​ y-coordinate of -1/radical3

(Direction- Type ordered pair)

1/√3 is the tangent of 30º or π/6 ... this is the reference angle

-1/√3 is the tangent of the reference angle in quadrants II and IV

so the x-values are
... 5π/6, 11π/6, and 17π/6

To find the points on the graph of y = tan(x) on the interval [π/2, 3π] that have a y-coordinate of -1/√3, we need to solve for x.

The equation y = tan(x) can be rewritten as tan(x) = -1/√3.

Let's find the values of x that satisfy this equation step by step:

Step 1: Take the inverse tangent (arctan) of both sides to isolate x:
arctan(tan(x)) = arctan(-1/√3)

Step 2: Simplify the right side of the equation:
x = arctan(-1/√3)

Step 3: Use a calculator to find the angle whose tangent is -1/√3:
x ≈ 7π/6 (in radians)

Therefore, the point on the graph of y = tan(x) on the interval [π/2, 3π] with a y-coordinate of -1/√3 is (7π/6, -1/√3).

To find the points on the graph of y = tan(x) on the interval [π/2, 3π] that have a y-coordinate of -1/√3, we can use the definition of the tangent function.

The tangent function is defined as the ratio of the sine of an angle to the cosine of that angle. So, we have tan(x) = sin(x)/cos(x).

We want to find the values of x on the interval [π/2, 3π] where tan(x) = -1/√3.

First, let's find the angles whose tangent is -1/√3. We can do this by taking the arctangent (inverse tangent) of -1/√3. Using the arctangent function or a calculator, we find that arctan(-1/√3) ≈ -π/6.

So, we need to find the values of x on the interval [π/2, 3π] such that sin(x)/cos(x) = -1/√3.

Next, let's write this expression in terms of sine and cosine. We have sin(x)/cos(x) = -1/√3. Multiplying both sides by cos(x), we get sin(x) = -(1/√3)cos(x).

Now, we can use the trigonometric identity sin^2(x) + cos^2(x) = 1 to rewrite the equation. We have (sin(x))^2 + (cos(x))^2 = 1. Dividing both sides by (cos(x))^2, we get (sin(x))^2/(cos(x))^2 + (cos(x))^2/(cos(x))^2 = 1/(cos(x))^2.

Since (sin(x))^2/(cos(x))^2 is equal to tan(x)^2, the equation becomes tan(x)^2 + 1 = 1/(cos(x))^2.

Rearranging this equation gives tan(x)^2 = 1 - 1/(cos(x))^2.

Since tan(x) = -1/√3, we can substitute this value into the equation to get (-1/√3)^2 = 1 - 1/(cos(x))^2.

Simplifying, we get 1/3 = 1 - 1/(cos(x))^2.

Now, let's solve this equation for cos(x). Subtracting 1/3 from both sides gives 1 - 1/3 = 1/(cos(x))^2.

Simplifying, we have 2/3 = 1/(cos(x))^2.

Taking the reciprocal of both sides gives (cos(x))^2 = 3/2.

Now, let's find the values of x on the interval [π/2, 3π] that satisfy (cos(x))^2 = 3/2.

Taking the square root of both sides gives cos(x) = ±√(3/2).

We can now find the values of x by finding the inverse cosine (arccos) of ±√(3/2). Using a calculator, we find that arccos(√(3/2)) ≈ π/6 and arccos(-√(3/2)) ≈ 5π/6.

However, we need to find the values of x on the interval [π/2, 3π].

The solutions π/6 and 5π/6 fall outside of this interval.

But if we add 2π to π/6 and 5π/6, they will fall within the given interval.

So, the solutions for x on the interval [π/2, 3π] with a y-coordinate of -1/√3 are x = π/6 + 2π and x = 5π/6 + 2π.

Converting these angles to the form of the ordered pair (x, y), we have:

(π/6 + 2π, -1/√3)
(5π/6 + 2π, -1/√3)

Therefore, the points on the graph of y = tan(x) on the interval [π/2, 3π] with a y-coordinate of -1/√3 are (π/6 + 2π, -1/√3) and (5π/6 + 2π, -1/√3).