How do you graph to solve the equation on the interval $[-2pi,2pi]$ for $tanx=sqrt3$ ?

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Answer 1 · 2018-01-09T00:21:23

$x={-(5pi)/3, -(2pi)/3, pi/3, (4pi)/3}$ .

Since tangent is positive, we know that the angle comes from either QI or QIII.

Since $tan(x) = sqrt(3)$ we know that $x$ is a $pi/3$ angle.

Within $0<x<2pi$ , we know that the angles $pi/3$ and $4pi/3$ have tangents of $sqrt(3)$ .

Since we're solving on $-2pi<x<2pi$ , we can subtract $2pi$ from each value and stay in the necessary interval:

$pi/3-2pi=-(5pi)/3$
$4pi/3-2pi=-(2pi)/3$

So our answers are $x={-(5pi)/3, -(2pi)/3, pi/3, (4pi)/3}$ .

How do you graph to solve the equation on the interval [-2pi,2pi][-2pi,2pi] for tanx=sqrt3tanx=sqrt3?