Solving for Theta in the Interval 0<θ<360

[MorallyObtuse]

Homework Statement

How do I solve for theta here:

√13 Sin(θ-56.31)=1 in the interval 0< θ <360

2. The attempt at a solution
sin (θ - 56.31) = 1/√13
sin (θ - 56.31) = sin ( 16.102 )
=> θ - 56.31 = 16.102
Hence θ = 72.412
or θ - 56.31 = 180 - 16.102
θ - 56.31 = 163.898

 

[Mentallic]

Yes that's correct. You had one line left to go, why did you stop?

And this is assuming there is an interval for \theta such as 0&lt;\theta&lt;360^o because then you'll have an infinite number of solutions.

 

[nalkapo]

yes, there are an infinite number of solutions. Think about the periot of sine.
for example, f(x+2*pi)=f(x)
...f(30+(-360))=f(30)=f(30+360)=f(30+720)=...