Which of the following are solutions to the given equation on the interval [0,2π)?
√3 tan(3x)+1=0
Select all that are true. If there are none that are true, do no select any before submitting.
x=17π/18
x=23π/18
x=π/9
x=π/18
x=7π/9
x=2π/9
x=13π/18
x=35π/18
x=14π/9
x=19π18
check my answer please and thank you:
x=17π/18
x=23π/18
x=35π/18
Your answer is correct.
√3 tan ( 3 x ) = - 1
Subtract 1 to both sides
√3 tan ( 3 x ) = - 1
Divide both sides by √3
tan ( 3 x ) = - 1/ √3
Take the invert tangent of both sides.
3 x = - π / 6 + n π
Because period of tangent is 2π
n = ±1 , ± 2 , ± 3...
x = ( - π / 6 ) / 3 + n π / 3
x = - π / 18 + n π / 3
Interval [0,2π) means x > 0 , x ⩽ 2π
x = 0 is not part of interval, x = 2 π is part of interval.
The solutions are:
for n = 1
x = - π / 18 + π / 3 =
- π / 18 + 6 π / 18 = 5 π / 18
for n = 2
x = - π / 18 + 2 π / 3 =
- π / 18 + 12 π / 18 = 11 π / 18
for n = 3
x = - π / 18 + 3 π / 3 =
- π / 18 + 18 π / 18 = 17 π / 18
for n = 4
x = - π / 18 + 4 π / 3 =
- π / 18 + 24 π / 18 = 23 π / 18
for n = 5
x = - π / 18 + 5 π / 3 =
- π / 18 + 30 π / 18 = 29 π / 18
for n = 6
x = - π / 18 + 6 π / 3 =
- π / 18 + 36 π / 18 = 35 π / 18
So:
x = 17 π / 18
x = 23 π / 18
x = 35 π / 18
I make typo in first line.
It should be written like this:
√3 tan ( 3 x ) + 1 = 0
Subtract 1 to both sides
√3 tan ( 3 x ) = - 1
Thank you !
To find the solutions to the equation √3 tan(3x) + 1 = 0 on the interval [0, 2π), you can follow these steps:
Step 1: Solve for tan(3x)
√3 tan(3x) + 1 = 0
√3 tan(3x) = -1
Step 2: Solve for tan(3x)
tan(3x) = -1/√3
Step 3: Find the reference angle
The reference angle is the acute angle between the terminal side of an angle and the x-axis. In this case, the reference angle is π/6 (30 degrees).
Step 4: Find the solutions within the interval [0, 2π)
Since tan(3x) is periodic with a period of π/3 (60 degrees), we can find the solutions within the given interval by adding multiples of π/3 to the reference angle.
The reference angle is π/6 (30 degrees), and adding multiples of π/3, we have:
x = π/6 + k(π/3)
Multiples of π/3 within the interval [0, 2π) are:
k = 0: x = π/6
k = 1: x = π/6 + π/3 = π/2
k = 2: x = π/6 + 2(π/3) = 5π/6
k = 3: x = π/6 + 3(π/3) = 3π/2
k = 4: x = π/6 + 4(π/3) = 7π/6
So, within the interval [0, 2π), the solutions are:
x = π/6, π/2, 5π/6, 3π/2, 7π/6
Let's check the given options:
x = 17π/18 -> outside the interval [0, 2π)
x = 23π/18 -> outside the interval [0, 2π)
x = π/9 -> not a solution to the equation
x = π/18 -> not a solution to the equation
x = 7π/9 -> not a solution to the equation
x = 2π/9 -> outside the interval [0, 2π)
x = 13π/18 -> not a solution to the equation
x = 35π/18 -> outside the interval [0, 2π)
x = 14π/9 -> not a solution to the equation
x = 19π/18 -> outside the interval [0, 2π)
Therefore, the correct solutions within the interval [0, 2π) are:
x = π/6, π/2, 5π/6, 3π/2, 7π/6
So, your chosen options x = 17π/18, x = 23π/18, and x = 35π/18 are not solutions within the given interval.