Solve the equation?

Question

Solve the equation?

Solve

$tan x - \sqrt{3} = 0$ $tanx-sqrt3=0$
,
$x$ $x$ $\in$ $in$ $(- π, - \frac{π}{2}) \cup (- \frac{π}{2}, \frac{π}{2}) \cup (\frac{π}{2}, π)$ $(-pi,-pi/2)uu(-pi/2,pi/2)uu(pi/2,pi)$

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Answer 1 · 2018-04-09T00:33:56

$x = \frac{π}{3}$ $x=pi/3$ or $x = - \frac{2 π}{3}$ $x=-(2pi)/3$

Explanation:

$tan (x) - \sqrt{3} = 0$ $tan(x)-sqrt(3)=0$
$XXX \to tan (x) = \sqrt{3}$ $color(white)("XXX")rarr tan(x)=sqrt(3)$

In Quadrant I, this is one of the standard triangles:
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Using the CAST notation for the Quadrants, a reference angle in Quadrant III will have the same $tan (x)$ $tan(x)$ value i.e. $(- π + \frac{π}{3})$ $(-pi+pi/3)$ will have the same value.
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Answer 2 · 2018-04-09T00:44:59

$x = \frac{π}{3} + k π$ $x = pi/3 + kpi$

Explanation:

$tan x = \sqrt{3}$ $tan x = sqrt3$
Trig table and unit circle give 2 solutions:
$x = \frac{π}{3}$ $x = pi/3$ and $x = \frac{π}{3} + π = \frac{4 π}{3}$ $x = pi/3 + pi = (4pi)/3$
General answer:
$x = \frac{π}{3} + k π$ $x = pi/3 + kpi$
Inside the interval $(- π, - \frac{π}{2})$ $(-pi, -pi/2)$ , the answer is $\frac{4 π}{3}$ $(4pi)/3$
Inside the interval $(- \frac{π}{2}, \frac{π}{2})$ $(- pi/2, pi/2)$ , the answer is $(\frac{π}{3})$ $(pi/3)$
No answer in the interval $(\frac{π}{2}, π)$ $(pi/2, pi)$

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Solve the equation?

Solve

$tan x - \sqrt{3} = 0$ $tanx-sqrt3=0$
,
$x$ $x$ $\in$ $in$ $(- π, - \frac{π}{2}) \cup (- \frac{π}{2}, \frac{π}{2}) \cup (\frac{π}{2}, π)$ $(-pi,-pi/2)uu(-pi/2,pi/2)uu(pi/2,pi)$

2 Answers

Explanation:

Explanation:

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Solve the equation?

Solve tanx−√3=0tanx-sqrt3=0 , xx∈in(−π,−π2)∪(−π2,π2)∪(π2,π)(-pi,-pi/2)uu(-pi/2,pi/2)uu(pi/2,pi)

2 Answers

Explanation:

Explanation:

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Solve

$tan x - \sqrt{3} = 0$ $tanx-sqrt3=0$
,
$x$ $x$ $\in$ $in$ $(- π, - \frac{π}{2}) \cup (- \frac{π}{2}, \frac{π}{2}) \cup (\frac{π}{2}, π)$ $(-pi,-pi/2)uu(-pi/2,pi/2)uu(pi/2,pi)$