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Question Number 195263 by dimentri last updated on 28/Jul/23

{ ((y sin x+cos x = 2)),((4sin x−2y cos x = y)) :} tan x =?

$$\:\:\:\:\:\:\begin{cases}{{y}\:\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\:=\:\mathrm{2}}\\{\mathrm{4sin}\:{x}−\mathrm{2}{y}\:\mathrm{cos}\:{x}\:=\:{y}}\end{cases} \\ $$$$\:\:\:\:\:\:\:\mathrm{tan}\:{x}\:=?\: \\ $$

Answered by cortano12 last updated on 28/Jul/23

{ ( ),( ) :} determinant (( )) { ((4y sin x + 4 cos x = 8)),((4y sin x −2y^2 cos x = y^2 )) :} (4+2y^2 ) cos x = 8−y^2 determinant ((( ((8−y^2 )/(2y^2 +4))))) { ((2y^2 sin x +2y cos x = 4y)),((4 sin x −2y cos x = y)) :} (2y^2 +4) sin x = 5y determinant ((( ((5y)/(2y^2 +4))))) determinant ((( ((5y)/(2y^2 +4)) . ((2y^2 +4)/(8−y^2 ))=((5y)/(8−y^2 )))))

$$\:\:\:\:\:\:\begin{cases}{ }\\{ }\end{cases} \\ $$$$ \\ $$$$ \begin{array}{|c|}{ }\\\hline\end{array} \\ $$$$\:\:\:\:\:\:\:\begin{cases}{\mathrm{4}{y}\:\mathrm{sin}\:{x}\:+\:\mathrm{4}\:\mathrm{cos}\:{x}\:=\:\mathrm{8}}\\{\mathrm{4}{y}\:\mathrm{sin}\:{x}\:−\mathrm{2}{y}^{\mathrm{2}} \:\mathrm{cos}\:{x}\:=\:{y}^{\mathrm{2}} }\end{cases} \\ $$$$\:\:\:\:\:\:\:\:\left(\mathrm{4}+\mathrm{2}{y}^{\mathrm{2}} \right)\:\mathrm{cos}\:{x}\:=\:\mathrm{8}−{y}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\begin{array}{|c|}{ \frac{\mathrm{8}−{y}^{\mathrm{2}} }{\mathrm{2}{y}^{\mathrm{2}} +\mathrm{4}}}\\\hline\end{array} \\ $$$$\:\:\:\:\:\:\:\:\:\begin{cases}{\mathrm{2}{y}^{\mathrm{2}} \:\mathrm{sin}\:{x}\:+\mathrm{2}{y}\:\mathrm{cos}\:{x}\:=\:\mathrm{4}{y}}\\{\mathrm{4}\:\mathrm{sin}\:{x}\:−\mathrm{2}{y}\:\mathrm{cos}\:{x}\:=\:{y}}\end{cases} \\ $$$$\:\:\:\:\:\:\:\:\:\:\left(\mathrm{2}{y}^{\mathrm{2}} +\mathrm{4}\right)\:\mathrm{sin}\:{x}\:=\:\mathrm{5}{y} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\begin{array}{|c|}{ \frac{\mathrm{5}{y}}{\mathrm{2}{y}^{\mathrm{2}} +\mathrm{4}}}\\\hline\end{array} \\ $$$$\:\:\:\:\:\:\:\:\begin{array}{|c|}{ \frac{\mathrm{5}{y}}{\mathrm{2}{y}^{\mathrm{2}} +\mathrm{4}}\:.\:\frac{\mathrm{2}{y}^{\mathrm{2}} +\mathrm{4}}{\mathrm{8}−{y}^{\mathrm{2}} }=\frac{\mathrm{5}{y}}{\mathrm{8}−{y}^{\mathrm{2}} }}\\\hline\end{array} \\ $$

Answered by Frix last updated on 28/Jul/23

{ ((y=((2−cos x)/(sin x)))),((y=((4sin x)/(1+2cos x)))) :} ((2−cos x)/(sin x))=((4sin x)/(1+2cos x)) cos^2 x +(3/2)cos x −1=0 cos x =(1/2) [∨cos x =−2] tan x =±((√(1−cos^2 x))/(cos x))=±(√3)

$$\begin{cases}{{y}=\frac{\mathrm{2}−\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}}\\{{y}=\frac{\mathrm{4sin}\:{x}}{\mathrm{1}+\mathrm{2cos}\:{x}}}\end{cases} \\ $$$$\frac{\mathrm{2}−\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}=\frac{\mathrm{4sin}\:{x}}{\mathrm{1}+\mathrm{2cos}\:{x}} \\ $$$$\mathrm{cos}^{\mathrm{2}} \:{x}\:+\frac{\mathrm{3}}{\mathrm{2}}\mathrm{cos}\:{x}\:−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{cos}\:{x}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\left[\vee\mathrm{cos}\:{x}\:=−\mathrm{2}\right] \\ $$$$\mathrm{tan}\:{x}\:=\pm\frac{\sqrt{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:{x}}}{\mathrm{cos}\:{x}}=\pm\sqrt{\mathrm{3}} \\ $$

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