Solve tan^2x-3operatorname{gx}+2=0

$$\tan ^ { 2 } x - 3 \operatorname { gx } + 2 = 0$$

Solve for g (complex solution)

$$g=\frac{\cos(2x)+3}{3x\left(\cos(2x)+1\right)}$$

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

Steps for Solving Linear Equation

Subtract \left(\tan(x)\right)^{2} from both sides. Anything subtracted from zero gives its negation.

$$-3gx+2=-\left(\tan(x)\right)^{2}$$

Subtract 2 from both sides.

$$-3gx=-\left(\tan(x)\right)^{2}-2$$

The equation is in standard form.

$$\left(-3x\right)g=-\left(\tan(x)\right)^{2}-2$$

Divide both sides by -3x.

$$\frac{\left(-3x\right)g}{-3x}=-\frac{\frac{1}{\left(\cos(x)\right)^{2}}+1}{-3x}$$

Dividing by -3x undoes the multiplication by -3x.

$$g=-\frac{\frac{1}{\left(\cos(x)\right)^{2}}+1}{-3x}$$

Divide -\left(\frac{1}{\left(\cos(x)\right)^{2}}+1\right) by -3x.

$$g=\frac{\frac{1}{\left(\cos(x)\right)^{2}}+1}{3x}$$

Solve for g

$$g=\frac{\left(\cos(x)\right)^{2}+1}{3x\left(\cos(x)\right)^{2}}$$

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

Steps for Solving Linear Equation

Subtract \left(\tan(x)\right)^{2} from both sides. Anything subtracted from zero gives its negation.

$$-3gx+2=-\left(\tan(x)\right)^{2}$$

Subtract 2 from both sides.

$$-3gx=-\left(\tan(x)\right)^{2}-2$$

The equation is in standard form.

$$\left(-3x\right)g=-\left(\tan(x)\right)^{2}-2$$

Divide both sides by -3x.

$$\frac{\left(-3x\right)g}{-3x}=-\frac{\frac{1}{\left(\cos(x)\right)^{2}}+1}{-3x}$$

Dividing by -3x undoes the multiplication by -3x.

$$g=-\frac{\frac{1}{\left(\cos(x)\right)^{2}}+1}{-3x}$$

Divide -\left(\frac{1}{\left(\cos(x)\right)^{2}}+1\right) by -3x.

$$g=\frac{\frac{1}{\left(\cos(x)\right)^{2}}+1}{3x}$$