Solve y=1/cos(z)-tan(z)x

$$y= \frac{ 1 }{ \cos ( z ) } - \tan ( z ) x$$

Solve for x

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

Steps for Solving Linear Equation

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

$$\frac{1}{\cos(z)}-\tan(z)x=y$$

Subtract \frac{1}{\cos(z)} from both sides.

$$-\tan(z)x=y-\frac{1}{\cos(z)}$$

The equation is in standard form.

$$\left(-\tan(z)\right)x=-\frac{1}{\cos(z)}+y$$

Divide both sides by -\tan(z).

$$\frac{\left(-\tan(z)\right)x}{-\tan(z)}=\frac{-\frac{1}{\cos(z)}+y}{-\tan(z)}$$

Dividing by -\tan(z) undoes the multiplication by -\tan(z).

$$x=\frac{-\frac{1}{\cos(z)}+y}{-\tan(z)}$$

Divide y-\frac{1}{\cos(z)} by -\tan(z).

$$x=-\frac{y\cos(z)-1}{\sin(z)}$$