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Question Number 102783 by bramlex last updated on 11/Jul/20

2(√3)−1 = 6sin (2θ−60^o )−2sin (2θ−30^o ) ⇒θ=?

$$\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}\:=\:\mathrm{6sin}\:\left(\mathrm{2}\theta−\mathrm{60}^{\mathrm{o}} \right)−\mathrm{2sin} \\ $$$$\left(\mathrm{2}\theta−\mathrm{30}^{\mathrm{o}} \right)\:\Rightarrow\theta=? \\ $$

Answered by 1549442205 last updated on 11/Jul/20

⇔6sin2θ×(1/2)−6cos2θ×((√3)/2)−(2sin2θ×((√3)/2)−2cos2θ×(1/2))=2(√3)−1 ⇔(3−(√3))sin2θ−(3(√3)−1)cos2θ=2(√3)−1(1) Put x=tanθ.We have (1)⇔(3−(√3))((2x)/(1+x^2 ))−(3(√3)−1)((1−x^2 )/(1+x^2 ))=2(√3)−1(2) 2(3−(√3))x−(3(√3)−1)(1−x^2 )=(2(√3)−1)(1+x^2 ) (6−2(√3))x−3(√3)+1+(3(√3)−1)x^2 −2(√3)+1−(2(√3)−1)x^2 =0 (√3)x^2 +2(3−(√3))x−(5(√3)−2)=0 tan𝛉=x_1 =1.361080303⇔𝛉≈53°41′41 tan𝛉=x_2 =−2.825181918⇔𝛉≈−70°30′30

$$\Leftrightarrow\mathrm{6sin2}\theta×\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{6cos2}\theta×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\left(\mathrm{2sin2}\theta×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\mathrm{2cos2}\theta×\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1} \\ $$$$\Leftrightarrow\left(\mathrm{3}−\sqrt{\mathrm{3}}\right)\mathrm{sin2}\theta−\left(\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{1}\right)\mathrm{cos2}\theta=\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}\left(\mathrm{1}\right) \\ $$$$\mathrm{Put}\:\mathrm{x}=\mathrm{tan}\theta.\mathrm{We}\:\mathrm{have} \\ $$$$\left(\mathrm{1}\right)\Leftrightarrow\left(\mathrm{3}−\sqrt{\mathrm{3}}\right)\frac{\mathrm{2x}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }−\left(\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{1}\right)\frac{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}\left(\mathrm{2}\right) \\ $$$$\mathrm{2}\left(\mathrm{3}−\sqrt{\mathrm{3}}\right)\mathrm{x}−\left(\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{1}\right)\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)=\left(\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}\right)\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right) \\ $$$$\left(\mathrm{6}−\mathrm{2}\sqrt{\mathrm{3}}\right)\mathrm{x}−\mathrm{3}\sqrt{\mathrm{3}}+\mathrm{1}+\left(\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{1}\right)\mathrm{x}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}−\left(\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}\right)\mathrm{x}^{\mathrm{2}} =\mathrm{0} \\ $$$$\sqrt{\mathrm{3}}\mathrm{x}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{3}−\sqrt{\mathrm{3}}\right)\mathrm{x}−\left(\mathrm{5}\sqrt{\mathrm{3}}−\mathrm{2}\right)=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{tan}\theta}=\boldsymbol{\mathrm{x}}_{\mathrm{1}} =\mathrm{1}.\mathrm{361080303}\Leftrightarrow\boldsymbol{\theta}\approx\mathrm{53}°\mathrm{41}'\mathrm{41} \\ $$$$\boldsymbol{\mathrm{tan}\theta}=\boldsymbol{\mathrm{x}}_{\mathrm{2}} =−\mathrm{2}.\mathrm{825181918}\Leftrightarrow\boldsymbol{\theta}\approx−\mathrm{70}°\mathrm{30}'\mathrm{30} \\ $$

Answered by bemath last updated on 11/Jul/20

Commented by floor(10²Eta[1]) last updated on 11/Jul/20

why the eq. has solution if that inequality happens?

$$\mathrm{why}\:\mathrm{the}\:\mathrm{eq}.\:\mathrm{has}\:\mathrm{solution}\:\mathrm{if}\:\mathrm{that} \\ $$$$\mathrm{inequality}\:\mathrm{happens}? \\ $$

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