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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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 $x = \frac{\pi}{3}+n\pi,\qquad n\text{ any integer.}$