Para resolver, podemos "abrir" o seno do lado direito e o cosseno do lado esquerdo: $$\sin{(a+b)} = \sin{a}\cos{b} + \sin{b}\cos{a}$$ $$\cos{(a+b)} = \cos{a}\cos{b} - \sin{b}\sin{a}$$ Assim, como $\sin{60} = \frac{\sqrt 3}{2}$ e $\cos{60} = \frac{1}{2}$ $$2\cdot \left(\sin{\theta}\cdot \frac{1}{2} - \cos{\theta}\cdot \frac{\sqrt 3}{2} \right) = \frac{1}{2}\cdot \cos{\theta} - \frac{\sqrt 3}{2}\cdot \sin{\theta}$$ dividindo tudo por $\cos{\theta}$ e sumindo com o denominador $2$ que divide os dois lados da equação: $$2\cdot \tan{\theta} - 2\cdot \sqrt{3} = 1 - \sqrt{3} \cdot \tan{\theta}$$ isolando $\tan{\theta}$ $$\tan{\theta} = \frac{1 + 2\sqrt 3}{2 + \sqrt 3} = -4 + 3\sqrt {3}$$ ou seja, $a$ e $b$ são inteiros $$Letra \ B$$

Utilizando o seno da diferença e o cosseno da soma temos que $2\sin(\theta - 60°) = 2(\sin(\theta)\cos(60°) - \cos(\theta)\sin(60°)) = 2(\sin(\theta)\dfrac{1}{2} -\cos(\theta)\dfrac{\sqrt{3}}{2}) $ $= \boxed{2\sin(\theta - 60°) = \sin(\theta) - \cos(\theta)\sqrt{3} }$ $cos(\theta + 60°) = \cos(\theta)\cos(60°) - \sin(\theta)\sin(60°) $ $=\boxed{cos(\theta + 60°) = \cos(\theta)\dfrac{1}{2} - \sin(\theta)\dfrac{\sqrt{3}}{2}}$ $\therefore$ $2\sin(\theta - 60°) = \cos(\theta + 60°)$ $=\sin(\theta) - \cos(\theta)\sqrt{3} = \cos(\theta)\dfrac{1}{2} - \sin(\theta)\dfrac{\sqrt{3}}{2}$ $\implies \sin(\theta) + \sin(\theta)\dfrac{\sqrt{3}}{2} = \cos(\theta)\dfrac{1}{2} +\cos(\theta)\sqrt{3}$ $\dfrac{2\sin(\theta) + \sin(\theta)\sqrt{3}}{2} = \dfrac{\cos(\theta) + \cos(\theta)2\sqrt{3}}{2}$ $\implies 2\sin(\theta) + \sin(\theta)\sqrt{3}= \cos(\theta) + \cos(\theta)2\sqrt{3}$ $=\sin(\theta)(2\ + \sqrt{3})= \cos(\theta)(1 + 2\sqrt{3})$ $\implies \dfrac{\sin(\theta)}{\cos(\theta)} = \tan(\theta) = \dfrac{1 + 2\sqrt{3}}{2\ + \sqrt{3}}$ $ = \boxed{\tan(\theta) = 3\sqrt{3} - 4} $ $\therefore$ $\boxed{a = -4} $ e $\boxed{b = 3}$ Portanto , $a$ e $b$ são números inteiros. $\textbf{Resposta : Letra B}$