Is there an actual textbook or online resource that has a tutorial to solve $a\sin x+b\cos x=c$ for $a, b, c$ being either positive or negative?
I tried to find these types of equations/functions in some trigonometry textbooks in my library, but they didn't mention it may be it is in a chapter that I don't know or it is presented with other words/title, I don't know.
Thanks
On
The usual method consists in writing that $a, b$ are the coordinates of a point in the plane, and use its polar coordinates, that is, $a=r\cos\theta$ and $b=r\sin\theta$. Then your equation becomes
$$r\cos\theta\sin x+r\sin\theta\cos x=c$$ $$\sin(\theta+ x)=\frac cr$$
If $\left|\frac cr\right|\leq 1$, then the solutions are
$$x=\arcsin\left(\frac cr\right)-\theta+2k\pi$$ and $$x=\pi-\arcsin\left(\frac cr\right)+\theta+2k\pi$$
for $k\in\Bbb Z$.
Equations like this can be solved by turning them into quadratic equations. For example, you can write $x=2y$, so that $$ \begin{align} \sin x &= 2\sin y\cos y\\ \cos x &= \cos^2 y - \sin^2 y \end{align} $$ the equation can then be written as $$ 2a \sin y\cos y + b(\cos^2 y - \sin^2 y) = c(\sin^2 y + \cos^2 y) $$ Dividing by $\cos^2y$, we get $$ (b+c)\tan^2 y - 2a\tan y + (c-b) = 0,\tag{1} $$ which is an ordinary quadratic equation. Alternatively, you can square the original equation: $$ a^2\sin^2 x + b^2\cos^2 x + 2ab\sin x\cos x = c^2(\sin^2 x + \cos^2 x) $$ Dividing by $\cos^2x$, we get $$ (a^2-c^2) \tan^2 x + 2ab\tan x + (b^2-c^2) = 0.\tag{2} $$ Both procedures will work; the first method is better though, because equation (2) has 4 solutions, and only 2 of them will also be solutions of the original equation, so you have to check them.