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This is a very basic question but that I need to know to solve a harder calculus question. How do I solve for $x$, for the problem $\tan(x) = \sqrt{3}$?

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    You can observe that it is $\pi/3$ radians ($60$ degrees). Just draw the usual $30$-$60$-$90$ triangle. If the htpotenuse is $1$, the sides are easy to write down. There are other solutions, one in the third quadrant, and you can then always change by a multiple of $2\pi$.2012-02-16
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    I forgot the 30-60-90 trick what is it again?2012-02-16
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    It's half an equilateral triangle, which tells you something about the relative lengths involved.2012-02-16
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    From the hypotenus, draw a line making a $60$ degree angle with the shorter of the two legs. By chasing angles, you can see this divides the original triangle into $2$ triangles, one equiateral and the other isosceles with base angles $30$. From this you can conclude that the line you just drew bisects the hypotenuse. so if hypotenuse is $1$, the short side is $1/2$, and by Pythagorean Theorem the other side is $\sqrt{3}/2$.2012-02-16
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    @Sean: Gerry Myerson is right, it is much cleaner to cut an equilateral into two halves.2012-02-16

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