How do you isolate $\theta$ from $\sin(\theta) - k\cos(\theta) = \frac{m_1}{m_2}$? This equation derives from physics force equations, and I have verified to make sure that the equation works. No actual physics pertains to this question. The part I need help on only requires math skills. Here is the equation:
$$\sin(\theta) - k\cos(\theta) = \frac{m_1}{m_2}$$
where $k$, $m_1$, and $m_2$ are variables, specifically the kinetic friction coefficient, mass one, and mass two respectively (irrelevant).
I have tried squaring both sides to then use double angle rules on the $2\sin\cos$ that yields, but all of my attempts have failed to simplify the equation further.
Could someone help me derive an equation for theta?
 A: Let $sin{\theta} = x$.
Then your equation becomes
$$x-k\sqrt{1-x^{2}} = \frac{m1}{m2}$$
Separate the linear terms and square both sides.
$$(x-\frac{m1}{m2})^{2} = k^{2}(1-x^{2})$$
It is now a simple quadratic equation which you can solve. But remember to choose only those solutions that lie within [-1,1] or can lie within [-1,1] for some values of $k,m1, m2$.
Now when you have obtained $\sin{\theta}$, use $\sin^{-1}$ to obtain $\theta$
A: It can be shown that any expression of the form $a\cos(\theta)+b\sin(\theta)$ ($a$ and $b$ are constants) can be written in the form $m\cos(\theta+c)$, where $m$ and $c$ are constants. Specifically, by taking $m=\text{sgn}(a)\sqrt{a^2+b^2}$ and $c=-\tan^{-1}\left(\frac{b}{a}\right)$, where $\text{sgn}(a)$ denotes the sign of $a$, we can write
$$a\cos(\theta)+b\sin(\theta)=\text{sgn}(a)\sqrt{a^2+b^2}\cos\left(\theta-\tan^{-1}\frac{b}{a}\right)$$
Substituting $a=-k$ and $b=1$ gives
\begin{align*}
\sin(\theta)-k\cos(\theta) &= \text{sgn}(-k)\sqrt{(-k)^2+1^2}\cos\left(\theta-\tan^{-1}\frac{1}{-k}\right)\\
&=-\text{sgn}(k)\sqrt{1+k^2}\cos\left(\theta+\tan^{-1}\frac{1}{k}\right)
\end{align*}
so if $k>0$, your equation boils down to
$$-\sqrt{1+k^2}\cos\left(\theta+\cot^{-1}k\right)=\frac{m_1}{m_2}$$
Dividing both sides by $-\sqrt{1+k^2}$ gives $\cos\left(\theta+\cot^{-1}k\right)=-\frac{m_1}{m_2\sqrt{1+k^2}}$, provided that $\left|-\frac{m_1}{m_2\sqrt{1+k^2}}\right|\leq 1$.
It can be shown that for any real number $y$ with $|y|\leq 1$, the only solutions to the equation $\cos(x)=y$ are of the form $x=2\pi m\pm\arccos(y)$ for integer $m$, so
$$\theta+\cot^{-1}(k)=2\pi m\pm\arccos\left(-\frac{m_1}{m_2\sqrt{1+k^2}}\right)$$
The desired result immediately follows.
$$\theta=2\pi m-\cot^{-1}(k)\pm\arccos\left(-\frac{m_1}{m_2\sqrt{1+k^2}}\right)$$
A: You might want to post the physics question which led you to this equation. It is likely there was a mistake in the derivation. Nonetheless,
$$\sin \theta - k \cos \theta = m_1/m_2 = \gamma$$
$$\sin \theta - k\sqrt{1-\sin^2 \theta} = \gamma$$
$$\sin \theta - \gamma = k \sqrt{1-\sin^2 \theta}$$
$$(\sin \theta -\gamma)^2 = k^2(1-\sin^2 \theta)$$
$$\sin^2 \theta - 2\sin \theta \gamma + \gamma^2 = k^2-k^2 \sin^2 \theta$$
$$(k^2+1) \sin^2 \theta  +(-2 \gamma) \sin \theta + \gamma^2 -k^2=0$$
$$\sin \theta = \frac{2\gamma \pm \sqrt{4\gamma^2 - 4(k^2+1)(\gamma^2 -k^2)}}{2(k^2+1)}$$
A: Let:
$$k=\tan \alpha \ \iff \ \alpha:=\tan^{-1}k.$$
The equation
$$\sin \theta - k \cos \theta = \underbrace{m_1/m_2}_{\gamma}$$
becomes, by multiplication by $\cos \alpha$:
$$\cos \alpha \sin \theta - \sin \alpha \cos \theta =\gamma \cos \alpha$$
$$\sin(\theta - \alpha)=\gamma \cos \alpha \tag{1}$$
If $\gamma \cos \alpha \in [-1,1]$, we can set
$$\sin \delta:= \gamma \cos \alpha \ \iff \ \delta=\sin^{-1}(\gamma \cos \alpha)$$
Equation (1) becomes:
$$\sin(\theta - \alpha)=\sin \delta \tag{2}$$
$$\iff \begin{cases}\theta - \alpha&=&\delta+k 2 \pi\\
\theta - \alpha&=&\pi-\delta+k' 2 \pi\end{cases}$$
whence two expressions for $\theta$.
A: Here's a general guide and explanation for problems of your type:
If we have an expression, $a\sin{x}+b\cos{x}$, let us assume it can be written in the form $R\sin(x+\alpha)$ Now to see if we can find values for $R$ and $\alpha$ in terms of $a$ and $b$. Using the compund angle formulae, also known as the addition formulae:
$$R\sin(x+\alpha)=R\sin{x}\cos{\alpha}+R\sin\alpha\cos x=a\sin{x}+b\cos{x}$$
So we have
$$R\cos\alpha=a,R\sin\alpha=b$$
So dividing the second equality by the first:
$$\tan\alpha=\frac{b}{a}$$
meaning we can find $\alpha$ in terms of $a$ and $b$, as we wanted. Now, to find $R$:
Squaring the $2$ equalities above we have
$$R^2\cos^2\alpha+R^2\sin^2\alpha=R^2(\cos^2\alpha+\sin^2\alpha)=R^2=a^2+b^2\implies R=\sqrt{a^2+b^2}$$
So, to finish off by recapping what we have learnt:
$$\tan\alpha=\frac{b}{a},R=\sqrt{a^2+b^2}$$
meaning
$$a\sin{x}+b\cos{x}\equiv\sqrt{a^2+b^2}\sin(\theta+\arctan{\frac{b}{a}})$$
Try applying that to your question. If you have any questions, please ask!
I hope that was helpful :)
Note: It's not always going to be the case that we take the positive square root for $R$. It depends on the sign of the quantities $a$ and $\cos\alpha$ ( I could've picked $b$ and $\sin\alpha$ as well; can you see why?).
