solution to  equation $a \cdot \cos(\theta) - b \cdot \sin(\theta) = c$ Does the equation
$$ a \cdot \cos(\theta) - b \cdot \sin(\theta) = c$$
have a closed-form solution for $\theta$?  What about the case where $a^2 + b^2 = 1$?

Given your use of signs I'll take it that $a,b \ge 0$. In which case, if $R = \sqrt{a^2+b^2}$ and $\alpha = \tan^{-1} \dfrac{b}{a}$ then
$$a \cos \theta - b\sin \theta = R\cos(\theta + \alpha)$$
and so if $\left| c \right| \le R$, we can solve the equation to get
$$\theta = \cos^{-1} \dfrac{c}{R} - \alpha = \boxed{\cos^{-1} \dfrac{c}{\sqrt{a^2+b^2}} - \tan^{-1} \dfrac{b}{a}}$$
Of course, this is just one solution; we can obtain the others by considering symmetries.
