Solve c=asin(x)+bcos(x)

$$c = a \sin ( x ) + b \cos ( x )$$

Solve for a

$$\left\{\begin{matrix}a=-\frac{b\cos(x)-c}{\sin(x)}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}\\a\in \mathrm{R}\text{, }&c=b\cos(x)\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}\end{matrix}\right.$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$a\sin(x)+b\cos(x)=c$$

Subtract b\cos(x) from both sides.

$$a\sin(x)=c-b\cos(x)$$

The equation is in standard form.

$$\sin(x)a=-b\cos(x)+c$$

Divide both sides by \sin(x).

$$\frac{\sin(x)a}{\sin(x)}=\frac{-b\cos(x)+c}{\sin(x)}$$

Dividing by \sin(x) undoes the multiplication by \sin(x).

$$a=\frac{-b\cos(x)+c}{\sin(x)}$$

Solve for b

$$\left\{\begin{matrix}b=-\frac{a\sin(x)-c}{\cos(x)}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}+\frac{\pi }{2}\\b\in \mathrm{R}\text{, }&c=a\sin(x)\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }x=\frac{\pi \left(2n_{1}+1\right)}{2}\end{matrix}\right.$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$a\sin(x)+b\cos(x)=c$$

Subtract a\sin(x) from both sides.

$$b\cos(x)=c-a\sin(x)$$

The equation is in standard form.

$$\cos(x)b=-a\sin(x)+c$$

Divide both sides by \cos(x).

$$\frac{\cos(x)b}{\cos(x)}=\frac{-a\sin(x)+c}{\cos(x)}$$

Dividing by \cos(x) undoes the multiplication by \cos(x).

$$b=\frac{-a\sin(x)+c}{\cos(x)}$$