Question

How do you solve `2 sin 2 θ = 1`?

Answer 1

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