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Question Number 43192 by Raj Singh last updated on 08/Sep/18

Commented by Rauny last updated on 09/Sep/18

$$\mathrm{cosec}x+\sec x=\frac{1}{\sin x}+\frac{1}{\cos x}=2$$$$\cos x+\sin x=2\sin x\cos x$$$$A=B\Rightarrow A^2=B^2,$$$${\cos }^{2}x+2\sin x\cos x+{\sin }^{2}x={4\sin }^{2}x{\cos }^{2}x$$$$\mathrm{by}\mathrm{the}\mathrm{way}\mathrm{\forall }x:{\sin }^{2}x+{\cos }^{2}x=1$$$$viz.2\sin x\cos x+1={4\sin }^{2}x{\cos }^{2}x$$$$\mathrm{let}\sin x\cos x=X,$$$$4X^2-2X-1=0$$$$X=\frac{1\pm \sqrt{5}}{4}=\sin x\cos x$$$${\sin }^{2}x{\cos }^{2}x=\frac{3\pm \sqrt{5}}{8}$$$$={\sin }^{2}x(1-{\sin }^{2}x)\because {\cos }^{2}x=1-{\sin }^{2}x$$$$-{\sin }^{4}x+{\sin }^{2}x-\frac{3\pm \sqrt{5}}{8}=0$$$$\mathrm{let}{\sin }^{2}x=S,$$$$S^2-S+\frac{3\pm \sqrt{5}}{8}=-0=0$$$$S=\frac{1+\sqrt{\frac{5\mp \sqrt{5}}{8}}}{2}\mathrm{or}\frac{1-\sqrt{\frac{5\mp \sqrt{5}}{8}}}{2}$$$$=\frac{1}{2}(1+\sqrt{\frac{5+\sqrt{5}}{8}})\mathrm{or}\frac{1}{2}(1+\sqrt{\frac{5-\sqrt{5}}{8}})$$$$\mathrm{or}\frac{1}{2}(1-\sqrt{\frac{5+\sqrt{5}}{8}})\mathrm{or}\frac{1}{2}(1-\sqrt{\frac{5-\sqrt{5}}{8}})$$$$\therefore \mathbf{cosec}\mathit{x}=\frac{2}{1+\sqrt{\frac{5+\sqrt{5}}{8}}}\mathrm{or}\frac{2}{1+\sqrt{\frac{5-\sqrt{5}}{8}}}$$$$\mathrm{or}\frac{2}{1-\sqrt{\frac{5+\sqrt{5}}{8}}}\mathrm{or}\frac{2}{1-\sqrt{\frac{5-\sqrt{5}}{8}}}$$$$\mathrm{or}\frac{-2}{1+\sqrt{\frac{5+\sqrt{5}}{8}}}\mathrm{or}\frac{-2}{1+\sqrt{\frac{5-\sqrt{5}}{8}}}$$$$\mathrm{or}\frac{-2}{1-\sqrt{\frac{5+\sqrt{5}}{8}}}\mathrm{or}\frac{-2}{1-\sqrt{\frac{5-\sqrt{5}}{8}}}$$$$◼$$$$\mathrm{omg}..$$

Commented by Rauny last updated on 10/Sep/18

$$$$$$\mathrm{I}\mathrm{made}\mathrm{so}\mathrm{many}\mathrm{mistakes}..$$$$therealsollution:\mathrm{cosec}x+\sec x=2$$$$\Rightarrow \sin x=\sqrt{\frac{1}{2}+\frac{\sqrt{5+2\sqrt{5}}}{2}i},$$$$\mathrm{cosec}x=\frac{(1-\sqrt{5+2\sqrt{5}})\sqrt{\frac{1}{2}+\frac{\sqrt{5+2\sqrt{5}}}{2}i}}{3+\sqrt{5}}$$$$=\frac{1}{4}(3-\sqrt{5})(1-\sqrt{5+2\sqrt{5}})\sqrt{\frac{1}{2}+\frac{\sqrt{5+2\sqrt{5}}}{2}i}$$$$\mathrm{I}\mathrm{have}\mathrm{to}\mathrm{sleep}\mathrm{now}.\mathrm{sorry}!$$

Commented by Rauny last updated on 14/Sep/18

$$REALLYEXCELLENTSOLLUTION:$$$$\mathrm{cosec}x+\sec x=2$$$$\sin x+\cos x=2\sin x\cos x$$$$\sin x=\frac{e^{ix}-e^{-ix}}{2i},\cos x=\frac{e^{ix}+e^{-ix}}{2}$$$$\mathrm{And}{e}^{ix}:=E$$$$\mathrm{Then}\frac{E-E^{-1}}{2i}+\frac{E+E^{-1}}{2}=\frac{(E-E^{-1})(E+E^{-1})}{2i}$$$$\Rightarrow \frac{E-E^{-1}+iE+{iE}^{-1}}{2i}=\frac{E^2-E^{-2}}{2i}$$$$(1+i)E-(1-i)E^{-1}=E^2-E^{-2}$$$$E^2-\frac{1}{E^2}-(1+i)E+\frac{1-i}{E}=0$$$$E^2+2i-\frac{1}{E^2}-E-Ei+\frac{1}{E}-\frac{1}{E}i-2i=0$$$${(E+\frac{1}{E}i)}^2-(E+\frac{1}{E}i)-(Ei-\frac{1}{E})-2i=0$$$$\Rightarrow {(E+\frac{1}{E}i)}^2-(E+\frac{1}{E}i)-(Ei+\frac{1}{E}{i}^2)-2i=0$$$$\Rightarrow {(E+\frac{1}{E}i)}^2-(E+\frac{1}{E}i)-(E+\frac{1}{E}i)i-2i=0$$$$E+\frac{1}{E}i:=X,$$$$X^2-X-Xi-2i=0$$$$X^2-(1+i)X-2i=0$$$$X=\frac{1\pm \sqrt{10i}+i}{2}=E+\frac{1}{E}i$$$$E^2+i=\frac{1\pm \sqrt{10i}+i}{2}E$$$$E^2-\frac{1\pm \sqrt{10i}+i}{2}E+i=0$$$$E=\frac{1}{2}(-\frac{1\pm \sqrt{10i}+i}{2}+\sqrt{\frac{\pm (1+\sqrt{10i}+i)\sqrt{10i}}{4}-\frac{7}{2}i}$$$$\mathrm{or}\frac{1}{2}(-\frac{1\pm \sqrt{10i}+i}{2}-\sqrt{\frac{\pm (1+\sqrt{10i}+i)\sqrt{10i}}{4}-\frac{7}{2}i}$$$$=e^{ix}$$$$1+i:=j,\sqrt{10i}:=k$$$$e^{ix}=\frac{1}{2}(-\frac{j\pm k}{2}+\sqrt{\pm \frac{j+k}{4}k-\frac{7}{2}i})$$$$\mathrm{or}\frac{1}{2}(-\frac{j\pm k}{2}-\sqrt{\pm \frac{j+k}{4}k-\frac{7}{2}i})$$$$\mathrm{Then}\sin x=\frac{e^{ix}-e^{-ix}}{2i}$$$$=\frac{1}{2i}[-\frac{j\pm k}{4}+\frac{1}{2}\sqrt{\pm \frac{j+k}{4}k-\frac{7}{2}i}-\{-(j\pm k)+2\sqrt{\pm \frac{j+k}{4}k-\frac{7}{2}i}\}]$$$$\mathrm{or}\frac{1}{2i}[-\frac{j\pm k}{4}-\frac{1}{2}\sqrt{\pm \frac{j+k}{4}k-\frac{7}{2}i}-\{-(j\pm k)-2\sqrt{\pm \frac{j+k}{4}k-\frac{7}{2}i}\}]$$$$=\frac{1}{2i}\{\frac{3}{4}(j\pm k)-\frac{3}{4}\sqrt{\pm (j+k)k-14i}\}$$$$\mathrm{or}\frac{1}{2i}\{\frac{3}{4}(j\pm k)+\frac{3}{4}\sqrt{\pm (j+k)k-14i}\}$$$$\mathrm{cosec}x=\frac{1}{\sin x}$$$$\therefore \mathrm{cosec}x+\sec x=2\wedge 1+i=j\wedge \sqrt{10i}=k$$$$\Rightarrow \mathrm{cosec}x=\frac{3i}{2}\{(j\pm k)+\sqrt{\pm (j+k)k-14i}\}$$$$\mathrm{or}\frac{3i}{2}\{(j\pm k)-\sqrt{\pm (j+k)k-14i}\}$$$$\mathrm{Q}.\mathrm{E}.\mathrm{D}.$$

Commented by Rauny last updated on 13/Sep/18

$$\mathrm{p}.\mathrm{s}.\mathrm{I}\mathrm{wrote}\mathrm{that}\mathrm{for}50\mathrm{minutes}.\mathrm{lol}$$

Commented by rahul 19 last updated on 13/Sep/18

My God! ����������

Commented by Rauny last updated on 14/Sep/18

$$:)$$

Answered by alex041103 last updated on 08/Sep/18

$$letcosec\varphi =t=\frac{1}{sin\varphi }>1$$$$\Rightarrow sec\varphi =\frac{1}{cos\varphi }=\frac{1}{\sqrt{1-\frac{1}{t^2}}}=\frac{t}{\sqrt{t^2-1}}$$$$\Rightarrow t+\frac{t}{\sqrt{t^2-1}}=2$$$$\frac{t}{\sqrt{t^2-1}}=2-t$$$$\frac{t^2}{t^2-1}=4+t^2-4t$$$$ift\ne 1(t=1isnotasolution)$$$$\Rightarrow t^2=(t^2-4t+4)(t^2-1)$$$$=t^4-4t^3+4t^2-t^2+4t-4$$$$\Rightarrow t^4-4t^3+2t^2+4t-4=0$$$$\Rightarrow therealsolutionsare:$$$$t_1=1-\sqrt{2+\sqrt{5}}$$$$t_2=1+\sqrt{2+\sqrt{5}}$$$$Thetwosolutionsare{t}_{1,2}\in \mathbb{R}-(-1;1)$$$$\Rightarrow {\varphi }_{1;2}exists$$$$\Rightarrow cosec\varphi =1-\sqrt{2+\sqrt{5}};1+\sqrt{2+\sqrt{5}}$$

Commented by malwaan last updated on 09/Sep/18

$$\mathrm{thank}\mathrm{you}$$

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