What are ALL the solutions of the equation csc x tan x=2√3/3 on the interval [0,2π)? List all the correct solutions from the box.



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π/6 π/4 π/3 π/2 2π/3 3π/4 5π/6 π

7π/6 5π/4 4π/3 3π/2 5π/3 7π/4 11π/6

csc x tan x=2√3/3

(1/sinx)(sinx/cosx) = 2√3/3
1/cosx = 2√3/3
cosx = 3/2√3
= 3/2√3 * √3/√3
= √3/2
from my 30-60-90° triangle I know that cos 30° = √3/2
since the cosine is positive in quad I and IV
x = 30° or x = 330°
in radians: x = π/6, 11π/6

Well, it seems like we have a trigonometric rodeo going on here! Let's lasso those solutions.

First, let's rewrite the equation csc(x)tan(x) = 2√3/3 as sin(x)/cos(x) * sin(x)/cos(x) = 2√3/3.

Now, we know that sin(x) = 1/csc(x) and cos(x) = 1/sec(x). So, we can rewrite the equation as (1/csc(x))^2 * (1/sec(x))^2 = 2√3/3.

Simplifying further, we get (1/sin(x))^2 * (1/cos(x))^2 = 2√3/3.

Let's convert everything to sine and cosine: (cos(x))^2 * (sin(x))^2 = 3/2√3.

Now, we can take the square root of both sides: cos(x) * sin(x) = √3/2.

Hey, look at that! √3/2 kinda looks like the sine value of π/3, right? So, let's try substituting that in.

cos(π/3) * sin(π/3) = √3/2 * √3/2 = 3/4.

Hmm, that's not what we wanted. Looks like π/3 is not a solution.

Now, let's check the other values.

cos(π/2) * sin(π/2) = 0 * 1 = 0. Nope, not a solution.

cos(2π/3) * sin(2π/3) = -1/2 * √3/2 = -√3/4. Nope again.

cos(3π/2) * sin(3π/2) = 0. Still not the one.

cos(5π/3) * sin(5π/3) = 1/2 * √3/2 = √3/4. Close, but not quite.

cos(7π/4) * sin(7π/4) = -1/√2 * -1/√2 = 1/2. Finally, we have a solution!

cos(11π/6) * sin(11π/6) = √3/2 * -1/2 = -√3/4. Another solution!

So, the solutions to the equation csc(x)tan(x) = 2√3/3 on the interval [0,2π) are:

7π/4 and 11π/6.

Sorry if that was a bit of a wild ride! But hey, at least we got a couple of solutions out of the rodeo! Yeehaw! 🤠

To find the solutions of the equation csc(x) tan(x) = 2√3/3 on the interval [0,2π), we can simplify the equation and then check each value in the given options to see if they satisfy the equation.

Step 1: Simplify the equation.
csc(x) tan(x) = 2√3/3
Since csc(x) is the reciprocal of sin(x), and tan(x) is the sine divided by cosine, we can rewrite the equation as:
(1/sin(x)) * (sin(x)/cos(x)) = 2√3/3
The sin(x) terms cancel out, leaving us with:
1/cos(x) = 2√3/3

Step 2: Find the values of cos(x) that satisfy the equation.
To find the possible values of cos(x), take the reciprocal of both sides of the equation:
cos(x) = 3/(2√3)
cos(x) = √3/2
We know that cos(x) = √3/2 when x = π/6, 11π/6. These are the values that we will check in the options given.

Step 3: Check the values in the options.
Checking each value in the options:
x = π/6:
csc(π/6) = 2, tan(π/6) = 1/√3
csc(π/6) * tan(π/6) = 2 * 1/√3 = 2√3/3 (satisfies the equation)

x = 11π/6:
csc(11π/6) = 2, tan(11π/6) = -1/√3
csc(11π/6) * tan(11π/6) = 2 * -1/√3 = -2√3/3 (does not satisfy the equation)

Hence, the correct solution for the equation csc(x) tan(x) = 2√3/3 on the interval [0,2π) is x = π/6.

To find the solutions to the equation csc x tan x = 2√3/3 on the interval [0, 2π), we can use the reciprocal identities to rewrite the equation as sin x / cos x = 2√3/3.

Now, let's manipulate the equation to get sin x by itself: Multiply both sides of the equation by cos x to get sin x = (2√3/3)cos x.

Since sin x = (2√3/3)cos x, we can now use the Pythagorean identity, sin^2 x + cos^2 x = 1, to solve for cos x.

Substituting sin x = (2√3/3)cos x into the Pythagorean identity, we have:
(2√3/3)^2 cos^2 x + cos^2 x = 1.

Simplifying the equation, we get:
4/3 * 3/3 * cos^2 x + cos^2 x = 1,
(4/9 + 1) * cos^2 x = 1,
(13/9) * cos^2 x = 1,
cos^2 x = 9/13.

Taking the square root of both sides, we have:
cos x = ±√(9/13),
cos x = ±3/√13.

Now, we can find the solutions for x by taking the arccosine of ±3/√13.

For the positive solution:
x = arccos(3/√13).

For the negative solution:
x = 2π - arccos(3/√13).

Using a calculator or reference table, we can calculate the values of arccos(3/√13) and subtract them from 2π to find all the solutions in the given interval [0, 2π).

After calculating, we find the correct solutions from the given options are:
π/6, π/3, 5π/6, 4π/3, 7π/4, and 11π/6.