Riješi 1=2(10rsin2θsinθcosphi)

$$1 = 2 ( 10 r \sin 2 \theta \sin \theta \cos \phi )$$

Riješite za r (complex solution)

$$r=-\frac{2e^{8i\theta }}{5\left(e^{iϕ+5i\theta }+e^{iϕ+11i\theta }-e^{iϕ+7i\theta }-e^{iϕ+9i\theta }+e^{5i\theta -iϕ}+e^{11i\theta -iϕ}-e^{7i\theta -iϕ}-e^{9i\theta -iϕ}\right)}$$

$$5e^{iϕ+7i\theta }+5e^{iϕ+9i\theta }-5e^{iϕ+5i\theta }-5e^{iϕ+11i\theta }+5e^{7i\theta -iϕ}+5e^{9i\theta -iϕ}-5e^{5i\theta -iϕ}-5e^{11i\theta -iϕ}\neq 0\text{ and }e^{iϕ+5i\theta }+e^{iϕ+11i\theta }-e^{iϕ+7i\theta }-e^{iϕ+9i\theta }+e^{5i\theta -iϕ}+e^{11i\theta -iϕ}-e^{7i\theta -iϕ}-e^{9i\theta -iϕ}\neq 0\text{ and }e^{-8i\theta }\left(e^{iϕ+5i\theta }+e^{iϕ+11i\theta }-e^{iϕ+7i\theta }-e^{iϕ+9i\theta }+e^{5i\theta -iϕ}+e^{11i\theta -iϕ}-e^{7i\theta -iϕ}-e^{9i\theta -iϕ}\right)\neq 0\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }ϕ=\pi n_{1}+\frac{\pi }{2}$$

Koraci za rješavanje linearne jednačine

Pomnožite 2 i 10 da biste dobili 20.

$$1=20r\sin(2\theta )\sin(\theta )\cos(ϕ)$$

Zamijenite strane tako da svi promjenljivi izrazi budu na lijevoj strani.

$$20r\sin(2\theta )\sin(\theta )\cos(ϕ)=1$$

Jednačina je u standardnom obliku.

$$20\sin(\theta )\sin(2\theta )\cos(ϕ)r=1$$

Podijelite obje strane s 20\sin(2\theta )\sin(\theta )\cos(ϕ).

$$\frac{20\sin(\theta )\sin(2\theta )\cos(ϕ)r}{20\sin(\theta )\sin(2\theta )\cos(ϕ)}=\frac{1}{20\sin(\theta )\sin(2\theta )\cos(ϕ)}$$

Dijelјenje sa 20\sin(2\theta )\sin(\theta )\cos(ϕ) poništava množenje sa 20\sin(2\theta )\sin(\theta )\cos(ϕ).

$$r=\frac{1}{20\sin(\theta )\sin(2\theta )\cos(ϕ)}$$

Podijelite 1 sa 20\sin(2\theta )\sin(\theta )\cos(ϕ).

$$r=\frac{1}{40\cos(\theta )\cos(ϕ)\left(\sin(\theta )\right)^{2}}$$

Riješite za r

$$r=\frac{1}{40\cos(\theta )\cos(ϕ)\left(\sin(\theta )\right)^{2}}$$

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

Koraci za rješavanje linearne jednačine

Pomnožite 2 i 10 da biste dobili 20.

$$1=20r\sin(2\theta )\sin(\theta )\cos(ϕ)$$

Zamijenite strane tako da svi promjenljivi izrazi budu na lijevoj strani.

$$20r\sin(2\theta )\sin(\theta )\cos(ϕ)=1$$

Jednačina je u standardnom obliku.

$$20\sin(\theta )\sin(2\theta )\cos(ϕ)r=1$$

Podijelite obje strane s 20\sin(2\theta )\sin(\theta )\cos(ϕ).

$$\frac{20\sin(\theta )\sin(2\theta )\cos(ϕ)r}{20\sin(\theta )\sin(2\theta )\cos(ϕ)}=\frac{1}{20\sin(\theta )\sin(2\theta )\cos(ϕ)}$$

Dijelјenje sa 20\sin(2\theta )\sin(\theta )\cos(ϕ) poništava množenje sa 20\sin(2\theta )\sin(\theta )\cos(ϕ).

$$r=\frac{1}{20\sin(\theta )\sin(2\theta )\cos(ϕ)}$$

Podijelite 1 sa 20\sin(2\theta )\sin(\theta )\cos(ϕ).

$$r=\frac{1}{40\cos(\theta )\cos(ϕ)\left(\sin(\theta )\right)^{2}}$$