Help with Trigonometry Equations [URGENT]

[animetony]

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Hey guys I need urgent help so please reply asap
I was studying for my SAC and came across this unusual question.
Find x, 0(degrees)<x<360(degrees). {The arrows have the underline as well}

tan^2x + (sqrt(3) - 1)tanx = sqrt(3)

HELP PLEASE

Thanks in advance :}

P.S The square root only applies to the 3

[dmpiq]

1. First, let a = tan(x)
2. Next, substitute the a into the equation: a^2 + (sqrt(3)-1)a = sqrt(3)
3. Transpose the sqrt(3) to the LHS: a^2 + (sqrt(3)-1)a-sqrt(3) = 0
4. Expand the brackets for the coefficient of a: a^2 + sqrt(3)a-a-sqrt(3) = 0
5. Rearrange the equation to be of the form a^2 - a + sqrt(3)a -sqrt(3) = 0.
6. Factorise by taking a common factor of (a-1) out of the first two and the last two terms: a(a-1)+ sqrt(3)(a-1) = 0
7. Now the equation becomes (a-1)(a+sqrt(3))=0
8. Using the null factor law: a-1 = 0 OR a+sqrt(3) = 0
9. a = 1 OR a = -sqrt(3)
10. Substitute a = tan(x) back. Hence, tan(x) = 1 or tan(x) = -sqrt(3)
11. x = pie/4, 5pie/4 OR x = 2pie/3, 5pie/3
12. Therefore, x = pie/4, 2pie/3, 5pie/4, 5pie/3 (all in radians)

[animetony]

Thanks for your kind help! I just asked my teacher today about this problem and he showed me how to do it. Thank you anyways for your generous help. Really Appreciated.

Sincerely yours, Animetony

[Cobalt]

12. Therefore, x = pie/4, 2pie/3, 5pie/4, 5pie/3 (all in radians)

I lol'd

Methods is a piece of pie.