Solve a/dxcos(x)=-sin(x)

$$\frac { a } { d x } \cos ( x ) = - \sin ( x )$$

Solve for a

$$a=-dx\tan(x)$$

$$d\neq 0\text{ and }x\neq 0\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}+\frac{\pi }{2}$$

Steps for Solving Linear Equation

Multiply both sides of the equation by dx.

$$a\cos(x)=-dx\sin(x)$$

The equation is in standard form.

$$\cos(x)a=-dx\sin(x)$$

Divide both sides by \cos(x).

$$\frac{\cos(x)a}{\cos(x)}=-\frac{dx\sin(x)}{\cos(x)}$$

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

$$a=-\frac{dx\sin(x)}{\cos(x)}$$

Divide -dx\sin(x) by \cos(x).

$$a=-dx\tan(x)$$

Solve for d (complex solution)

$$\left\{\begin{matrix}d=-\frac{a\cot(x)}{x}\text{, }&\nexists n_{2}\in \mathrm{Z}\text{ : }x=\frac{\pi n_{2}}{2}\text{ and }a\neq 0\\d\neq 0\text{, }&\exists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}\text{ and }a=0\text{ and }x\neq 0\end{matrix}\right.$$

Solve for d

$$\left\{\begin{matrix}d=-\frac{a\cot(x)}{x}\text{, }&\exists n_{2}\in \mathrm{Z}\text{ : }\left(x>\frac{\pi n_{2}}{2}\text{ and }x<\frac{\pi n_{2}}{2}+\frac{\pi }{2}\right)\text{ and }a\neq 0\\d\neq 0\text{, }&\exists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}\text{ and }a=0\text{ and }x\neq 0\end{matrix}\right.$$