$$\cos(\theta-60^\circ)=\frac{1}{2}\sin(\theta)$$ $$2\cos(\theta - 60^\circ)=\sin(\theta)$$ $$2=\frac{\sin(\theta)}{\cos(\theta-60^\circ)}$$
How do I get rid of the $60^\circ$ so that I can make this a solvable trig identity?
Edit: This is my whole solution so you can see where I am going wrong: $$\cos(\theta-60^\circ)=\frac{1}{2}\sin(\theta)$$ $$2\cos(\theta)\cos(60^\circ)+2\sin(\theta)\sin(60^\circ)=\sin(\theta)$$ $$\frac{2\cos(\theta)+2\sin(\theta)(\sqrt3)}{2}=\sin(\theta)$$ $$\cos(\theta)+\sin(\theta)(\sqrt3)=\sin(\theta)$$ $$\frac{\cos(\theta)}{\sin(\theta)}+\sqrt3=\frac{\sin(\theta)}{\sin(\theta)}$$ $$\cot(\theta)+\sqrt3=1$$ $$\cot(\theta)=1-\sqrt3$$ $$\theta=\cot^{-1}(1-\sqrt3)$$
Thanks in advance