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

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

I-solve ang c (complex solution)

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

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

Mga Hakbang sa Pagso-solve ng Linear Equation

Pagpalitin ang magkabilang panig para nasa kaliwang bahagi ang lahat ng variable na term.

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

I-subtract ang \tan(x) mula sa magkabilang dulo.

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

Ang equation ay nasa standard form.

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

I-divide ang magkabilang dulo ng equation gamit ang 2.

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

Kapag na-divide gamit ang 2, ma-a-undo ang multiplication gamit ang 2.

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

I-divide ang 2\cot(x) gamit ang 2.

$$c=\cot(x)$$

I-solve ang c

$$c=\cot(x)$$

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

Mga Hakbang sa Pagso-solve ng Linear Equation

Pagpalitin ang magkabilang panig para nasa kaliwang bahagi ang lahat ng variable na term.

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

I-subtract ang \tan(x) mula sa magkabilang dulo.

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

Ang equation ay nasa standard form.

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

I-divide ang magkabilang dulo ng equation gamit ang 2.

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

Kapag na-divide gamit ang 2, ma-a-undo ang multiplication gamit ang 2.

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

I-divide ang 2\cot(x) gamit ang 2.

$$c=\cot(x)$$

I-solve ang x

$$x=2\pi n_{13}+\left(-1\right)\pi +ArcCosI(c\left(1+c^{2}\right)^{-\frac{1}{2}})\text{, }n_{13}\in \mathrm{Z}\text{, }\exists n_{6}\in \mathrm{Z}\text{ : }\left(n_{13}>\left(-\frac{1}{2}\right)\left(\left(-1\right)\pi +ArcCosI(c\left(1+c^{2}\right)^{-\frac{1}{2}})+\left(-\frac{1}{2}\right)\pi n_{6}\right)\pi ^{-1}\text{ and }2\pi n_{13}+\left(-1\right)\pi +ArcCosI(c\left(1+c^{2}\right)^{-\frac{1}{2}})<\frac{1}{2}\pi \left(n_{6}+1\right)\right)\text{ and }\exists n_{6}\in \mathrm{Z}\text{ : }\left(2\pi n_{13}+\left(-1\right)\pi +ArcCosI(c\left(1+c^{2}\right)^{-\frac{1}{2}})>\frac{1}{2}\pi n_{6}\text{ and }2\pi n_{13}+\left(-1\right)\pi +ArcCosI(c\left(1+c^{2}\right)^{-\frac{1}{2}})<\frac{1}{2}\pi \left(n_{6}+1\right)\right)$$

$$x=ArcCosI(c\left(1+c^{2}\right)^{-\frac{1}{2}})+2\pi n_{351}\text{, }n_{351}\in \mathrm{Z}\text{, }\exists n_{6}\in \mathrm{Z}\text{ : }\left(ArcCosI(c\left(1+c^{2}\right)^{-\frac{1}{2}})+2\pi n_{351}>\frac{1}{2}\pi n_{6}\text{ and }ArcCosI(c\left(1+c^{2}\right)^{-\frac{1}{2}})+2\pi n_{351}<\frac{1}{2}\pi \left(n_{6}+1\right)\right)\text{ and }\exists n_{6}\in \mathrm{Z}\text{ : }\left(ArcCosI(c\left(1+c^{2}\right)^{-\frac{1}{2}})+2\pi n_{351}>\frac{1}{2}\pi n_{6}\text{ and }ArcCosI(c\left(1+c^{2}\right)^{-\frac{1}{2}})+2\pi n_{351}<\frac{1}{2}\pi \left(n_{6}+1\right)\right)$$