Resolver operatorname{cotg}(2x-pi)=operatorname{estg}(x+pi/3)

$$\operatorname { cotg } ( 2 x - \pi ) = \operatorname { estg } ( x + \frac { \pi } { 3 } )$$

Ebatzi: s

$$\left\{\begin{matrix}s=-\frac{\cot(g)\left(2\sqrt{3}x\sin(x)-2x\cos(x)-\pi \sqrt{3}\sin(x)+\pi \cos(x)\right)}{e\left(\sqrt{3}\cos(x)+\sin(x)\right)}\text{, }&\nexists n_{4}\in \mathrm{Z}\text{ : }x=\frac{\pi n_{4}}{2}+\frac{\pi }{6}\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }g=\pi n_{1}\\s\in \mathrm{R}\text{, }&\exists n_{3}\in \mathrm{Z}\text{ : }g=\pi n_{3}+\frac{\pi }{2}\text{ and }\nexists n_{2}\in \mathrm{Z}\text{ : }x=\pi n_{2}+\frac{\pi }{6}\text{ and }\exists n_{3}\in \mathrm{Z}\text{ : }x=\pi n_{3}+\frac{2\pi }{3}\end{matrix}\right.$$

Ekuazio lineala ebazteko urratsak

Erabili banaketa-propietatea \cot(g) eta 2x-\pi biderkatzeko.

$$2\cot(g)x-\cot(g)\pi =es\tan(x+\frac{\pi }{3})$$

Trukatu aldeak, aldagaiak ezkerraldean egon daitezen.

$$es\tan(x+\frac{\pi }{3})=2\cot(g)x-\cot(g)\pi$$

Modu arruntean dago ekuazioa.

$$e\tan(x+\frac{\pi }{3})s=2x\cot(g)-\pi \cot(g)$$

Zatitu ekuazioaren bi aldeak e\tan(x+\frac{1}{3}\pi ) balioarekin.

$$\frac{e\tan(x+\frac{\pi }{3})s}{e\tan(x+\frac{\pi }{3})}=\frac{\left(2x-\pi \right)\cot(g)}{e\tan(x+\frac{\pi }{3})}$$

e\tan(x+\frac{1}{3}\pi ) balioarekin zatituz gero, e\tan(x+\frac{1}{3}\pi ) balioarekiko biderketa desegiten da.

$$s=\frac{\left(2x-\pi \right)\cot(g)}{e\tan(x+\frac{\pi }{3})}$$

Zatitu \left(2x-\pi \right)\cot(g) balioa e\tan(x+\frac{1}{3}\pi ) balioarekin.

$$s=\frac{\left(2x-\pi \right)\cot(g)\left(-\sqrt{3}\sin(x)+\cos(x)\right)}{2e\cos(\frac{6x-\pi }{6})}$$