Løs 3tanx=5K

$$3 \tan x = 5 K$$

Løs for K

$$K=\frac{3\tan(x)}{5}$$

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

Trinn for å løse lineære ligninger

Bytt om sidene, slik at alle variabelledd er på venstre side.

$$5K=3\tan(x)$$

Del begge sidene på 5.

$$\frac{5K}{5}=\frac{3\tan(x)}{5}$$

Hvis du deler på 5, gjør du om gangingen med 5.

$$K=\frac{3\tan(x)}{5}$$

Løs for x

$$x=\pi +2n_{3}\pi +arcSin(5K\left(25K^{2}+9\right)^{-\frac{1}{2}})\text{, }n_{3}\in \mathrm{Z}\text{, }\exists n_{50}\in \mathrm{Z}\text{ : }\left(n_{3}>\left(-\frac{1}{2}\right)\left(\frac{1}{2}\pi +arcSin(5K\left(25K^{2}+9\right)^{-\frac{1}{2}})+\left(-1\right)\pi n_{50}\right)\pi ^{-1}\text{ and }n_{3}<\left(-\frac{1}{2}\right)\left(\left(-\frac{1}{2}\right)\pi +arcSin(5K\left(25K^{2}+9\right)^{-\frac{1}{2}})+\left(-1\right)\pi n_{50}\right)\pi ^{-1}\right)$$

$$x=arcSin(5K\left(25K^{2}+9\right)^{-\frac{1}{2}})+2\pi n_{26}\text{, }n_{26}\in \mathrm{Z}\text{, }\exists n_{50}\in \mathrm{Z}\text{ : }\left(n_{50}<\pi ^{-1}\left(arcSin(5K\left(25K^{2}+9\right)^{-\frac{1}{2}})+2\pi n_{26}+\left(-\frac{1}{2}\right)\pi \right)\text{ and }n_{50}>\left(arcSin(5K\left(25K^{2}+9\right)^{-\frac{1}{2}})+2\pi n_{26}+\left(-\frac{3}{2}\right)\pi \right)\pi ^{-1}\right)$$