Is that correct, sqrt(3/3) ?

Why, sqrt(3/3) is sqrt(1). And sqrt(1) is +,-1.

So, arccot[sqrt(3/3)] = arccot(+,-1)

And so, arccot[-sqrt(3/3)] = arccot[-(+,-1)] = arccot(-,+1) = arccot(+,-1)

Cotangent (and Tangent) function is positive in the 1st and 3rd quadrants,

So, Arccot (1) is 45 or 225 degrees. Or, pi/4 and 5pi/4 radians.

Cotangent (and Tangent) function is negative in the 2nd and 4th quadrants,

So, Arccot (-1) is 135 or 315 degrees. Or, 3pi/4 and 7pi/4 radians.

Therefore, arccot[-sqrt(3/3)] = pi/4, 3pi/4, 5pi/4, or 7pi/4 radians in the interval [0,2pi]. -------answer.

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However, I doudt if that is really sqrt(3/3)....

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Addition....

I think that is arccot[- (1/3)sqrt(3)]

(1/3)sqrt(3) is actully 1/sqrt(3) if you rationalize the denominator.

Hence the reference angle is 60 degrees or pi/3 radians,

because cot(60deg) = 1/sqrt(3)

Then, arccot[-(1/3)sqrt(3)] -----in the 2nd or 4th quadrants,

= arccot[-1/sqrt(3)]

= 120 or 300 degrees

= 2pi/3 or 5pi/3 radians --------answer.