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How do you solve #2 sin 2 θ = 1#?

1 Answer

Konstantinos Michailidis

Sep 27, 2015

See explanation

Explanation:

We have that

#2sin2theta=1=>sin2theta=1/2=>sin2theta=sin(pi/6)#

so the solutions are

#2theta=2k*pi+pi/6=>theta=(12kpi+pi)/12#

and

#2theta=2kpi+(5pi)/6=>theta=(12kpi+5pi)/12#

where #k# belongs to the set of integers

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