arctan [ ( square root 3 ) / (3) ]
y = arctan [ (  square root 3) / (3) ]
tan(y) = [ ( square root 3) / (3)]
y = 5pi/6, 11pi/6 > ANSWER
Remember the unit circle in trig
Quad 1 all positive
Quad 2 sine is positive
Quad 3 tangent is positive
Quad 4 cosine is positive
Just by looking at the sign conventions you can eliminate quad 1 and quad 3. Your angles have to be in quad 2 and quad 4.
The angle has to be something n * (pi/6) because tan(pi/6) = 0.57 which is what we want but negative.
So the only ones are x = 5pi/6, and x = 11pi/6.
y = arctan [ (  square root 3) / (3) ]
tan(y) = [ ( square root 3) / (3)]
y = 5pi/6, 11pi/6 > ANSWER
Remember the unit circle in trig
Quad 1 all positive
Quad 2 sine is positive
Quad 3 tangent is positive
Quad 4 cosine is positive
Just by looking at the sign conventions you can eliminate quad 1 and quad 3. Your angles have to be in quad 2 and quad 4.
The angle has to be something n * (pi/6) because tan(pi/6) = 0.57 which is what we want but negative.
So the only ones are x = 5pi/6, and x = 11pi/6.

arctan(√3/3) = 5pi/6, 11pi/6