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Question Number 22701 by selestian last updated on 22/Oct/17

Answered by ajfour last updated on 22/Oct/17

e^((((sin^2 x)/(1−sin^2 x)))ln 2) =e^(ln 2^(tan^2 x) ) =2^(tan^2 x) roots of x^2 −9x+8=0 are 1, 8 and 2^(tan^2 x) also satisfies the eqn. So 2^(tan^2 x) =8, 1 ⇒ tan^2 x = 3, or 0 And if 0 < x < (π/2) , then tan x = (√3) ((cos x)/(cos x+sin x)) = (1/(1+tan x)) =(1/(1+(√3))) .

$${e}^{\left(\frac{\mathrm{sin}\:^{\mathrm{2}} {x}}{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} {x}}\right)\mathrm{ln}\:\mathrm{2}} \:={e}^{\mathrm{ln}\:\mathrm{2}^{\mathrm{tan}\:^{\mathrm{2}} {x}} } \:=\mathrm{2}^{\mathrm{tan}\:^{\mathrm{2}} {x}} \\ $$$${roots}\:{of}\:{x}^{\mathrm{2}} −\mathrm{9}{x}+\mathrm{8}=\mathrm{0}\:\:{are}\:\mathrm{1},\:\mathrm{8} \\ $$$${and}\:\mathrm{2}^{\mathrm{tan}\:^{\mathrm{2}} {x}} \:{also}\:{satisfies}\:{the}\:{eqn}. \\ $$$${So}\:\:\:\:\mathrm{2}^{\mathrm{tan}\:^{\mathrm{2}} {x}} \:=\mathrm{8},\:\mathrm{1} \\ $$$$\Rightarrow\:\:\mathrm{tan}\:^{\mathrm{2}} {x}\:=\:\:\mathrm{3},\:{or}\:\mathrm{0} \\ $$$${And}\:{if}\:\:\mathrm{0}\:<\:{x}\:<\:\frac{\pi}{\mathrm{2}}\:,\:{then} \\ $$$$\:\:\:\:\mathrm{tan}\:{x}\:=\:\sqrt{\mathrm{3}}\: \\ $$$$\frac{\mathrm{cos}\:{x}}{\mathrm{cos}\:{x}+\mathrm{sin}\:{x}}\:=\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{tan}\:{x}}\:=\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{3}}}\:. \\ $$

Answered by ajfour last updated on 22/Oct/17

x^2 −3x+2m=0 let its roots be 2α, β x^2 −x+m=0 let its roots be α, γ Then 2α+β=3 , 2αβ=2m Also α+β=1 , αβ=m (2α+β)−(α+β)=3−1 ⇒ α=2, β=−1 and m=−2 .

$${x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}{m}=\mathrm{0} \\ $$$${let}\:{its}\:{roots}\:{be}\:\mathrm{2}\alpha,\:\beta \\ $$$${x}^{\mathrm{2}} −{x}+{m}=\mathrm{0} \\ $$$${let}\:{its}\:{roots}\:{be}\:\alpha,\:\gamma \\ $$$${Then}\:\:\:\:\mathrm{2}\alpha+\beta=\mathrm{3}\:\:\:,\:\:\:\mathrm{2}\alpha\beta=\mathrm{2}{m} \\ $$$${Also}\:\:\:\:\:\:\:\:\alpha+\beta=\mathrm{1}\:\:,\:\:\:\:\:\alpha\beta={m} \\ $$$$\:\:\:\:\left(\mathrm{2}\alpha+\beta\right)−\left(\alpha+\beta\right)=\mathrm{3}−\mathrm{1} \\ $$$$\Rightarrow\:\:\:\alpha=\mathrm{2},\:\beta=−\mathrm{1}\:{and}\:\boldsymbol{{m}}=−\mathrm{2}\:. \\ $$

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