Assuming \(\theta\) is measured in degrees, let's first convert to radians.
\(\dfrac{60}2^\circ=30^\circ\times\dfrac{\pi\text{ rad}}{180^\circ}=\dfrac\pi6\)
So the equation is
\(\dfrac32=\dfrac{\sin\left(\frac\theta2+\frac\pi6\right)}{\sin\frac\theta2}\)
\(3\sin\dfrac\theta2=2\sin\left(\dfrac\theta2+\dfrac\pi6\right)\)
\(3\sin\dfrac\theta2=2\left(\sin\dfrac\theta2\cos\dfrac\pi6+\cos\dfrac\theta2\sin\dfrac\pi6\right)\)
\(3\sin\dfrac\theta2=\sqrt3\sin\dfrac\theta2+\cos\dfrac\theta2\)
\(3=\sqrt3+\cot\dfrac\theta2\)
\(3-\sqrt3=\cot\dfrac\theta2\)
\(\mathrm{arccot}(3-\sqrt3)+2n\pi=\dfrac\theta2\)
\(\theta=2\mathrm{arccot}(3-\sqrt3)+2n\pi\)
where \(n\) is any integer.
