<s>[INST] Trig - Difference formulas to solve $\sqrt3\sin \theta- \cos\theta = 1$ My exam review states that I need to utilize the difference formula for sine to solve the equation on the interval $0 \leq \theta < 2\pi $
$$\sqrt3\sin \theta- \cos\theta = 1$$
I know that: $\sin \frac\pi3 = \frac{\sqrt3}{2}$
and $\cos\frac\pi3 = \frac12 $, so I divide each term by 2 and rewrite the equation:
$$\frac{\sqrt3\sin\theta}{2} - \frac{\cos\theta}{2} = \frac12$$
from here, I apply the difference formula for sine:
$$\sin(\alpha-\beta) = (\sin\alpha \cdot \cos\beta)  - (\cos\alpha  \cdot \sin\beta)$$ 
(this is the step that I believe I'm doing incorrectly. I've tried plugging in the corresponding numerical values for sin/cos into the equation instead, but I was still unable to go past this step.)   
$$\sin(\alpha-\beta) = \left(\frac {\pi}{3} \cdot \frac{\pi}{3}\right)  - \left(\frac{\pi}{6} \cdot \frac{\pi}{6}\right) = \frac12$$ 
 [/INST] $$\sqrt3\sin \theta- \cos \theta = 1$$
$$\frac{\sqrt3}{2} \sin \theta- \frac{1}{2}\cos \theta = \frac{1}{2}$$
Now, $\sin \dfrac{\pi}{6}=\dfrac{ 1}{2}$ and $\cos \dfrac{\pi}{6}=\dfrac{ \sqrt{3}}{2}$
$$\cos \dfrac{\pi}{6} \sin \theta- \sin \dfrac{\pi}{6}\cos \theta = \frac{1}{2}$$
Let $\dfrac{\pi}{6}=A$
$$\sin(\theta -A) = \sin \theta \cos A - \cos \theta \sin A$$
$$\sin \Big(\theta - {\dfrac{\pi}{6}}\Big) =\cos \dfrac{\pi}{6} \sin \theta- \sin \dfrac{\pi}{6}\cos \theta = \frac{1}{2}$$
$$\implies  \Big(\theta - {\dfrac{\pi}{6}}\Big)= n\pi+(-1)^n\frac{\pi}{6}$$
$$\implies  \theta = \Big(n+\frac{1}{6} \Big)\pi+(-1)^n\frac{\pi}{6}$$
</s>