Solve the equation?

Solve

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

2 Answers
Alan P.
Apr 9, 2018

x=pi/3x=π3 or x=-(2pi)/3x=−2π3

Explanation:

tan(x)-sqrt(3)=0tan(x)−√3=0
color(white)("XXX")rarr tan(x)=sqrt(3)XXX→tan(x)=√3

In Quadrant I, this is one of the standard triangles:
enter image source here

Using the CAST notation for the Quadrants, a reference angle in Quadrant III will have the same tan(x)tan(x) value i.e. (-pi+pi/3)(−π+π3) will have the same value.
enter image source here

Nghi N
Apr 9, 2018

x = pi/3 + kpix=π3+kπ

Explanation:

tan x = sqrt3tanx=√3
Trig table and unit circle give 2 solutions:
x = pi/3x=π3 and x = pi/3 + pi = (4pi)/3x=π3+π=4π3
General answer:
x = pi/3 + kpix=π3+kπ
Inside the interval (-pi, -pi/2)(−π,−π2), the answer is (4pi)/34π3
Inside the interval (- pi/2, pi/2)(−π2,π2), the answer is (pi/3)(π3)
No answer in the interval (pi/2, pi)(π2,π)

Related questions
Impact of this question
1634 views around the world
You can reuse this answer
Creative Commons License