How do you solve the following equation #cot (x/2)=1# in the interval [0, 2pi]?

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How do you solve the following equation #cot (x/2)=1# in the interval [0, 2pi]?

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Answer 1 · 2016-09-11T12:00:26

#pi/2#

Explanation:

Trig table of special arcs and unit circle -->
#cot (x/2) = 1# --> #x/2 = pi/4 + kpi#
General answers:
#x = pi/2 + k2pi#
Answer in interval (0, 2pi) --> k = 0
#x = pi/2#

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How do you solve the following equation #cot (x/2)=1# in the interval [0, 2pi]?

1 Answer

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