Solving $1+\cos\alpha-R\sin\alpha=0$ So I was doing this physics problem, and I have simplified the problem to solve for $\alpha$. The problem is that I cannot solve it:

$$1+\cos\alpha-R\sin\alpha=0$$

($R$ is a dimensionless constant.)
My process so far was trying to isolate the function and condense it into one trig function to solve for $\alpha$. However, I don't have many steps and maybe I am approaching this the wrong way.
$$R\sin\alpha-\cos\alpha=1$$
Expanding $1$ with its Pythagorean identity we have
$$R\sin\alpha-\cos\alpha=\sin^2\alpha+\cos^2\alpha$$
Now I fear that I have messed up because instead of making it simpler I made it harder to solve for $\alpha$.
 A: Rearrange the equation as
$$1+\cos\alpha-R\sin\alpha$$
$$=2\cos^2\frac \alpha2-2R\cos\frac \alpha2\sin\frac \alpha2$$
$$=2\cos^2\frac \alpha2 \left(1 - R\tan\frac \alpha2\right)=0$$
which has two sets of solutions, given by 
$$\cos\frac \alpha2 = 0 \implies \alpha= \pi+2n\pi$$
$$\tan\frac \alpha2= \frac1R \implies 2\tan^{-1}\frac1R + 2n\pi$$
A: $R\sin\alpha - \cos \alpha = 1$
One solution is $\alpha = \pi$
$\sqrt{R^2 + 1}(\frac{R}{\sqrt{R^2+1}}\sin\alpha - \frac {1}{\sqrt{R^2+1}}\cos \alpha) = 1$
Let $\phi = \arctan R\\
\sin\phi = \frac {R}{\sqrt{R^2 + 1}}\\
\cos\phi = \frac {1}{\sqrt{R^2 + 1}}$
$\sin\phi\sin\alpha - \cos\phi\cos \alpha = \cos\phi\\
\cos{\alpha + \phi} = -\cos\phi\\
\alpha + \phi = \pi - \phi\\
\alpha  =  \pi - 2\phi\\
\alpha  = \pi- 2\arctan R$
gives another solution.
A: According to the identity $\sin(a-b) = \sin(a)\cos(b) - \sin(b)\cos(a)$, one obtains
\begin{align*}
& R\sin(\alpha) - \cos(\alpha) = 1 \Longleftrightarrow \frac{R\sin(\alpha)}{\sqrt{R^{2}+1}} - \frac{\cos(\alpha)}{\sqrt{R^{2}+1}} = \frac{1}{\sqrt{R^{2}+1}} \Longleftrightarrow\\\\
& \sin(\alpha -\theta) = \frac{1}{\sqrt{R^{2}+1}} \Longleftrightarrow \alpha = \arcsin\left(\frac{1}{\sqrt{R^{2}+1}}\right) + \theta = 2\theta
\end{align*}
where
\begin{align*}
\begin{cases}
\displaystyle\cos(\theta) = \frac{R}{\sqrt{R^{2}+1}}\\\\
\displaystyle\sin(\theta) = \frac{1}{\sqrt{R^{2}+1}}
\end{cases}
\end{align*}
A: Welcome to Maths SX!
The standard method rewrites your equation as
\begin{multline}\sqrt{R^2+1}\biggl(\frac R{\sqrt{R^2+1}}\sin\alpha-\frac 1{\sqrt{R^2+1}}\cos\alpha\biggr)=1\iff\\\frac R{\sqrt{R^2+1}}\sin\alpha-\frac 1{\sqrt{R^2+1}}\cos\alpha=\frac1{\sqrt{R^2+1}}\end{multline}
Now $\dfrac R{\sqrt{R^2+1}}=\cos \varphi,\quad \dfrac 1{\sqrt{R^2+1}}=\sin \varphi\;$ for a unique $\varphi\bmod 2\pi$,and once you've calculated $\varphi$, there remains to dolve the trigonometric equation
$$\sin(\alpha-\varphi)=\frac 1{\sqrt{R^2+1}}=\sin\varphi \iff\begin{cases}
\alpha-\varphi\equiv \varphi\mod 2\pi\\\alpha-\varphi\equiv \pi-\varphi\mod 2\pi\end{cases}\iff \begin{cases}
\alpha\equiv 2\varphi\mod 2\pi,\\\alpha\equiv \pi\mod 2\pi.\end{cases} $$
