Which of the following are solutions to the given equation on the interval [0,2π)?

Question

√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

Answer 1

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

Answer 2

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

Answer 3

Thank you !