Solve 1+operatorname{tes}4x=cos2x

$$1 + \operatorname { tes } 4 x = \cos 2 x$$

Solve for s_4 (complex solution)

$$\left\{\begin{matrix}s_{4}=-\frac{-\cos(2x)+1}{etx}\text{, }&x\neq 0\text{ and }t\neq 0\\s_{4}\in \mathrm{C}\text{, }&\exists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract 1 from both sides.

$$tes_{4}x=\cos(2x)-1$$

The equation is in standard form.

$$etxs_{4}=\cos(2x)-1$$

Divide both sides by tex.

$$\frac{etxs_{4}}{etx}=\frac{\cos(2x)-1}{etx}$$

Dividing by tex undoes the multiplication by tex.

$$s_{4}=\frac{\cos(2x)-1}{etx}$$

Solve for t (complex solution)

$$\left\{\begin{matrix}t=-\frac{-\cos(2x)+1}{es_{4}x}\text{, }&x\neq 0\text{ and }s_{4}\neq 0\\t\in \mathrm{C}\text{, }&\exists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract 1 from both sides.

$$tes_{4}x=\cos(2x)-1$$

The equation is in standard form.

$$es_{4}xt=\cos(2x)-1$$

Divide both sides by es_{4}x.

$$\frac{es_{4}xt}{es_{4}x}=\frac{\cos(2x)-1}{es_{4}x}$$

Dividing by es_{4}x undoes the multiplication by es_{4}x.

$$t=\frac{\cos(2x)-1}{es_{4}x}$$

Solve for s_4

$$\left\{\begin{matrix}s_{4}=-\frac{2\left(\sin(x)\right)^{2}}{etx}\text{, }&t\neq 0\text{ and }x\neq 0\\s_{4}\in \mathrm{R}\text{, }&\exists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}\text{, }not(n_{1}=0)\text{ and }t=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract 1 from both sides.

$$tes_{4}x=\cos(2x)-1$$

The equation is in standard form.

$$etxs_{4}=\cos(2x)-1$$

Divide both sides by tex.

$$\frac{etxs_{4}}{etx}=\frac{\cos(2x)-1}{etx}$$

Dividing by tex undoes the multiplication by tex.

$$s_{4}=\frac{\cos(2x)-1}{etx}$$

Solve for t

$$\left\{\begin{matrix}t=-\frac{2\left(\sin(x)\right)^{2}}{es_{4}x}\text{, }&x\neq 0\text{ and }s_{4}\neq 0\\t\in \mathrm{R}\text{, }&\left(\exists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}\text{ and }s_{4}=0\right)\text{ or }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract 1 from both sides.

$$tes_{4}x=\cos(2x)-1$$

The equation is in standard form.

$$es_{4}xt=\cos(2x)-1$$

Divide both sides by es_{4}x.

$$\frac{es_{4}xt}{es_{4}x}=\frac{\cos(2x)-1}{es_{4}x}$$

Dividing by es_{4}x undoes the multiplication by es_{4}x.

$$t=\frac{\cos(2x)-1}{es_{4}x}$$