Help with VCAA 2007 Trig Question

[Special At Specialist]

Right.
While we are discussing trig, how do I get the negative pi answer for this question? And why is that the case

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sqrt(3) sin(x) = cos(x) for x E [-pi, pi]
Divide both sides by cos(x) and since sin(x) / cos(x) = tan(x), we can write it as such:
sqrt(3) tan(x) = 1
tan(x) = 1 / sqrt(3)
Basic Angle = 30 degrees = pi/6
Tan is positive in quadrants 1 and 3, so the solutions between 0 and 2pi are:
x = pi/6 or x = pi + pi/6
x = pi/6 or x = 7pi/6
Since we only want the solutions between [-pi, pi], we can simply subtract 2pi from our answers:
x = pi/6 or x = 7pi/6 - 2pi
x = pi/6 or x = -5pi/6
x = -5pi/6 or x = pi/6 are the 2 solutions over the interval [-pi, pi].