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 $\tan(x) = \sqrt{3}$?
Basic Trig problem
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algebra-precalculus
trigonometry
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1You 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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0I forgot the 30-60-90 trick what is it again? – 2012-02-16
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0It's half an equilateral triangle, which tells you something about the relative lengths involved. – 2012-02-16
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0From 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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0@Sean: Gerry Myerson is right, it is much cleaner to cut an equilateral into two halves. – 2012-02-16
1 Answers
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You either use the $\arctan(y)$ function and plug in $\sqrt{3}$; or if you happen to know some of the basic values of the trigonometric functions, you might know that $$\sin(60^{\circ}) = \frac{\sqrt{3}}{2},\qquad\text{and}\qquad\cos(60^{\circ})=\frac{1}{2}$$ and therefore $$\tan(60^{\circ}) = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}.$$ In radians, $x=\frac{\pi}{3}$ is one solution.
Since the tangent function is periodic with period $180^{\circ}=\pi\ \text{radians}$, the solutions are $$x = \frac{\pi}{3}+n\pi,\qquad n\text{ any integer.}$$