Solve the equation?

Solve

#tanx-sqrt3=0#
,
#x##in##(-pi,-pi/2)uu(-pi/2,pi/2)uu(pi/2,pi)#

2 Answers
Alan P.
Apr 9, 2018

#x=pi/3# or #x=-(2pi)/3#

Explanation:

#tan(x)-sqrt(3)=0#
#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)# value i.e. #(-pi+pi/3)# will have the same value.
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Nghi N
Apr 9, 2018

#x = pi/3 + kpi#

Explanation:

#tan x = sqrt3#
Trig table and unit circle give 2 solutions:
#x = pi/3# and #x = pi/3 + pi = (4pi)/3#
General answer:
#x = pi/3 + kpi#
Inside the interval #(-pi, -pi/2)#, the answer is #(4pi)/3#
Inside the interval #(- pi/2, pi/2)#, the answer is #(pi/3)#
No answer in the interval #(pi/2, pi)#

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