Réitigh Qcosθ=frac{sqrt{2}}{2}

$$Q \cos \theta = \frac { \sqrt { 2 } } { 2 }$$

Réitigh do Q.

$$Q=\frac{\sqrt{2}}{2\cos(\theta )}$$

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

Céimeanna chun Cothromóidí Líneacha a Réiteach

Méadaigh an dá thaobh den chothromóid faoi 2.

$$2Q\cos(\theta )=\sqrt{2}$$

Tá an chothromóid i bhfoirm chaighdeánach.

$$2\cos(\theta )Q=\sqrt{2}$$

Roinn an dá thaobh faoi 2\cos(\theta ).

$$\frac{2\cos(\theta )Q}{2\cos(\theta )}=\frac{\sqrt{2}}{2\cos(\theta )}$$

Má roinntear é faoi 2\cos(\theta ) cuirtear an iolrúchán faoi 2\cos(\theta ) ar ceal.

$$Q=\frac{\sqrt{2}}{2\cos(\theta )}$$

Réitigh do θ. (complex solution)

$$\theta =2\pi n_{1}+\frac{-2i\ln(\frac{-\sqrt{1-2Q^{2}}+1}{Q})+\ln(2)i}{2}\text{, }n_{1}\in \mathrm{Z}$$

$$\theta =2\pi n_{2}+\frac{-2i\ln(\frac{\sqrt{1-2Q^{2}}+1}{Q})+\ln(2)i}{2}\text{, }n_{2}\in \mathrm{Z}\text{, }Q\neq 0$$

Réitigh do θ.

$$\theta =-\arccos(\frac{\sqrt{2}}{2Q})+2\pi n_{1}\text{, }n_{1}\in \mathrm{Z}$$

$$\theta =\arccos(\frac{\sqrt{2}}{2Q})+2\pi n_{2}\text{, }n_{2}\in \mathrm{Z}\text{, }|Q|\geq \frac{\sqrt{2}}{2}$$