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This question can be found in Heffernan's 2007 exam 1 for Maths Methods:

Solve over the domain [0,2*pi]:

tan2(x) + (1-rt(3))tan(x) = rt(3)

My working:
Let tan(x) = A

A2 + (1-rt(3))A = rt(3)
A2 + [(rt(3)/rt(3)) - (rt(3)/1)]A - rt(3) = 0
A2 + [(rt(3)-3rt(3))/rt(3)]A - rt(3) = 0

And that's where I become lost, did I simplify the cofactor 'A' correctly? How would I complete this square? Once I have completed this square, how would I solve the equation?

Cheers!

The fastest way to do this, although not intuitive or obvious is to go from step 1:

A2 + [1 - rt(3)]A - rt(3) = 0

Then factorise: 1 and -rt(3) add to [1 - rt(3)] and multiply to -rt(3)

Therefore: [A - rt(3)] * [A + 1] = 0

Hence, A = rt(3) and A = -1

Then go from there -- let me know if you need more help and I'll edit in a more comprehensive answer.

BTW, completing the square works if you didn't see the factorisation, but it will be rather painful. I'd recommend the quadratic formula (which does the same thing as complete the square BTW).

x = [ -b +/- rt(b2 - 4ac) ] / 2
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