Resolver 3tanx=5K

$$3 \tan x = 5 K$$

Ebatzi: K

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

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

Ekuazio lineala ebazteko urratsak

Trukatu aldeak, aldagaiak ezkerraldean egon daitezen.

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

Zatitu ekuazioaren bi aldeak 5 balioarekin.

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

5 balioarekin zatituz gero, 5 balioarekiko biderketa desegiten da.

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

Ebatzi: x

$$x=2\pi n_{1}+\arcsin(\frac{5K}{\sqrt{25K^{2}+9}})+\pi \text{, }n_{1}\in \mathrm{Z}\text{, }\exists n_{3}\in \mathrm{Z}\text{ : }\left(n_{1}>\frac{2n_{3}-\frac{2\arcsin(\frac{5K}{\sqrt{25K^{2}+9}})}{\pi }-1}{4}\text{ and }n_{1}<\frac{2n_{3}-\frac{2\arcsin(\frac{5K}{\sqrt{25K^{2}+9}})}{\pi }+1}{4}\right)$$

$$x=2\pi n_{2}+\arcsin(\frac{5K}{\sqrt{25K^{2}+9}})\text{, }n_{2}\in \mathrm{Z}\text{, }\exists n_{3}\in \mathrm{Z}\text{ : }\left(n_{3}>\frac{4n_{2}+\frac{2\arcsin(\frac{5K}{\sqrt{25K^{2}+9}})}{\pi }-3}{2}\text{ and }n_{3}<\frac{4n_{2}+\frac{2\arcsin(\frac{5K}{\sqrt{25K^{2}+9}})}{\pi }-1}{2}\right)$$