How do you graph to solve the equation on the interval $[- 2 π, 2 π]$ $[-2pi,2pi]$ for $tan x = \sqrt{3}$ $tanx=sqrt3$ ?

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Answer 1 · 2018-01-09T00:21:23

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

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

Since $tan (x) = \sqrt{3}$ $tan(x) = sqrt(3)$ we know that $x$ $x$ is a $\frac{π}{3}$ $pi/3$ angle.

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

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

$\frac{π}{3} - 2 π = - \frac{5 π}{3}$ $pi/3-2pi=-(5pi)/3$
$4 \frac{π}{3} - 2 π = - \frac{2 π}{3}$ $4pi/3-2pi=-(2pi)/3$

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

How do you graph to solve the equation on the interval [−2π,2π][-2pi,2pi] for tanx=√3tanx=sqrt3?