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.)   
$$\sin(\alpha-\beta) = \left(\frac {\pi}{3} \cdot \frac{\pi}{3}\right)  - \left(\frac{\pi}{6} \cdot \frac{\pi}{6}\right) = \frac12$$ 
 A: $$\sqrt3\sin \theta- \cos \theta = 1$$
$$2(\frac{\sqrt3}{2} \sin \theta- \frac{1}{2}\cos \theta)=1$$
$$(\frac{\sqrt3}{2} \sin \theta- \frac{1}{2}\cos \theta)=\frac12$$
$$ (\cos \dfrac{\pi}{6}\sin \theta -\sin \dfrac{\pi}{6} \cos \theta) = \sin \dfrac{\pi}{6} $$
$$\sin\left(\theta-\dfrac\pi6\right)=\sin\dfrac\pi6$$
$$\left(\theta-\dfrac\pi6\right)=\dfrac\pi6,\, \pi-\dfrac\pi6 .... $$
A: What you are missing is how to match the addition formula and the given equation.
You have
$$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta$$
vs.
$$\sin \theta\cdot\frac{\sqrt3}2-\cos\theta\cdot\frac12.$$
Then you identify
$$\alpha\leftrightarrow\theta\text{ and }\beta\leftrightarrow \frac\pi6$$
and get
$$\sin\left(\theta-\frac\pi6\right)=\frac12.$$
You should be able to finish.
A: $$\sin\left(\theta-\dfrac\pi6\right)=\sin\dfrac\pi6$$
Now $\sin x=\sin A,x=n\pi+(-1)^nA$ where $n$ is any integer
A: $$\sqrt3\sin \theta- \cos \theta = 1$$
$$\frac{\sqrt3}{2} \sin \theta- \frac{1}{2}\cos \theta = \frac{1}{2}$$
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 $\dfrac{\pi}{6}=A$
$$\sin(\theta -A) = \sin \theta \cos A - \cos \theta \sin A$$
$$\sin \Big(\theta - {\dfrac{\pi}{6}}\Big) =\cos \dfrac{\pi}{6} \sin \theta- \sin \dfrac{\pi}{6}\cos \theta = \frac{1}{2}$$
$$\implies  \Big(\theta - {\dfrac{\pi}{6}}\Big)= n\pi+(-1)^n\frac{\pi}{6}$$
$$\implies  \theta = \Big(n+\frac{1}{6} \Big)\pi+(-1)^n\frac{\pi}{6}$$
A: Approach this systematically as follows. Express your LHS (left-hand side) as
$R\sin(\theta-B)$ where you need to find $R$ and $B$. So
$$
R\sin(\theta-B)=R (\sin(\theta)\cos(B)-\cos(\theta)\sin(B))
$$
Comparing with $\sin(\theta)$ and $\cos(\theta)$ on your LHS we find
$$
R\cos(B)=\sqrt3
$$
$$
 R\sin(B)=1
$$
Now square and add these
$$
R^2=4 \Rightarrow R=2
$$
and divide these
$$
\tan{B}=\frac{1}{\sqrt3} \Rightarrow B=\pi/6
$$
Thus
$$
2\sin(\theta-\pi/6)=1 \Rightarrow \sin(\theta-\pi/6)=1/2
$$
Finally
$$
\theta-\pi/6 = n\pi+(-1)^n\pi/6
$$
