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Question Number 51600 by malwaan last updated on 28/Dec/18

$$\mathrm{solve}$$$$\left(\sin \theta \right){\mathrm{Z}}^2-\mathrm{i}\left(\cos \theta \right)\mathrm{Z}+\frac{1}{4}\sin \theta =0$$

Commented by Abdo msup. last updated on 29/Dec/18

$$\mathrm{\Delta }={(-icos\theta )}^2-4sin\theta \left(\frac{1}{4}sin\theta \right)=-{cos}^2\theta -{sin}^2\theta$$$$=-1\Rightarrow Z_1=\frac{icos\theta +i}{2sin\theta }=i\frac{2{cos}^2\left(\frac{\theta }{2}\right)}{4sin\left(\frac{\theta }{2}\right)cos\left(\frac{\theta }{2}\right)}$$$$=\frac{i}{2tan\left(\frac{\theta }{2}\right)}$$$$Z_2=\frac{icos\theta -i}{2sin\theta }=-i\frac{1-cos\theta }{2sin\theta }=-i\frac{2{sin}^2\left(\frac{\theta }{2}\right)}{4sin\left(\frac{\theta }{2}\right)cos\left(\frac{\theta }{2}\right)}$$$$=-\frac{i}{2}tan\left(\frac{\theta }{2}\right)(wesupposesin\theta \ne 0)$$$$ifsin\theta =0\Rightarrow cos\theta =\stackrel{-}{+}1and\left(e\right)\iff -icos\theta Z=0\Rightarrow$$$$Z=0.$$

Answered by Tawa1 last updated on 29/Dec/18

Commented by malwaan last updated on 29/Dec/18

$$\mathrm{thank}\mathrm{you}$$$$\mathrm{please}\mathrm{solve}\mathrm{it}\mathrm{with}\mathrm{this}\mathrm{method}$$$$\mathrm{Z}=\mathrm{x}+\mathrm{iy}$$

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