How to solve $A\sin{\theta} + B\cos{\theta} = C$? I've stumbled upon a equation in the form
$$A\sin{\theta} + B\cos{\theta} = C$$
What would be the steps necessary to solving it?
Thank you.
 A: Typical is to let $u = \arctan(A/B)$ (dealing with the special case of $B
 = 0$ separately), $r = \sqrt{A^2 + B^2}$. Then
\begin{align}
A &= r \sin u\\ 
B &= r \cos u
\end{align}
(unless I've swapped those two). Now your equation reads
$$
r \sin u \sin \theta + r \cos u \cos \theta = C
$$
which you rewrite as
\begin{align}
\cos(u-\theta) &= \frac{C}{r}\\
\theta &= u - \arccos(\frac{C}{r})
\end{align}
and you're done.
NB: If $|C| > |r|$, then there is no (real) solution. (Hat-tip to @LuisFelipe for making me add the "real".) If $r$ is zero and $C$ is nonzero, there's also no solution. If $r = C = 0$, then every value of $\theta$ is a solution. That leaves only the special case $B = 0, A \ne 0$ for you to work out.
A: $$A\sin{\theta} + B\cos{\theta} = C$$
use the identity
$$\cos\theta= \pm \sqrt{1-\sin^2\theta}$$
so
$$A\sin{\theta} \pm B\sqrt{1-\sin^2\theta} = C$$
$$ \pm B\sqrt{1-\sin^2\theta} = C-A\sin{\theta}$$
$$ B^2(1-\sin^2\theta) = (C-A\sin{\theta})^2$$
let $x=\sin{\theta}$
$$ B^2(1-x^2) = (C-Ax)^2$$
$$B^2-B^2x^2=C^2-2ACx+A^2x^2$$
$$(A^2+B^2)x^2-(2AC)x+(C^2-B^2)=0$$
then use the quadratic formula to solve it and then complete the solution to find the $\theta$ values
A: If either $A$ or $B$ equals zero we can use basic trigonometry to solve the equation, so let's assume neither $A$ nor $B$ is zero. Under this assumption we can write $$A\cos(\theta)+B\sin(\theta)=\alpha\cos\Big(\theta -\arctan(B/A)\Big)$$ where $$\alpha=A\cos(\arctan(B/A))+B\sin(\arctan(B/A))$$ Can you continue from here?
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:
$$\tan\alpha=\frac{B}{A},~R=\sqrt{A^2+B^2}$$
Try applying that to your question. I hope that was helpful :)
