Solve sin^2(x)+2cos^2(x)/sin(x)cos(x)=tan(x)+2c

$$\frac { \sin ^ { 2 } ( x ) + 2 \cos ^ { 2 } ( x ) } { \sin ( x ) \cos ( x ) } = \tan ( x ) + 2 c$$

Solve for c (complex solution)

$$c=\frac{\cos(2x)+1}{\sin(2x)}$$

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

Steps for Solving Linear Equation

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

$$\tan(x)+2c=\frac{\left(\sin(x)\right)^{2}+2\left(\cos(x)\right)^{2}}{\sin(x)\cos(x)}$$

Subtract \tan(x) from both sides.

$$2c=\frac{\left(\sin(x)\right)^{2}+2\left(\cos(x)\right)^{2}}{\sin(x)\cos(x)}-\tan(x)$$

The equation is in standard form.

$$2c=\frac{2\left(\cos(x)\right)^{2}+\left(\sin(x)\right)^{2}}{\frac{1}{2}\sin(2x)}-\tan(x)$$

Divide both sides by 2.

$$\frac{2c}{2}=\frac{2\cot(x)}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$c=\frac{2\cot(x)}{2}$$

Divide 2\cot(x) by 2.

$$c=\cot(x)$$

Solve for c

$$c=\cot(x)$$

$$\exists n_{1}\in \mathrm{Z}\text{ : }\left(x>\frac{\pi n_{1}}{2}\text{ and }x<\frac{\pi n_{1}}{2}+\frac{\pi }{2}\right)$$

Steps for Solving Linear Equation

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

$$\tan(x)+2c=\frac{\left(\sin(x)\right)^{2}+2\left(\cos(x)\right)^{2}}{\sin(x)\cos(x)}$$

Subtract \tan(x) from both sides.

$$2c=\frac{\left(\sin(x)\right)^{2}+2\left(\cos(x)\right)^{2}}{\sin(x)\cos(x)}-\tan(x)$$

The equation is in standard form.

$$2c=\frac{2\left(\cos(x)\right)^{2}+\left(\sin(x)\right)^{2}}{\frac{1}{2}\sin(2x)}-\tan(x)$$

Divide both sides by 2.

$$\frac{2c}{2}=\frac{2\cot(x)}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$c=\frac{2\cot(x)}{2}$$

Divide 2\cot(x) by 2.

$$c=\cot(x)$$