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Question-157080

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Question Number 157080 by amin96 last updated on 19/Oct/21
Commented by cortano last updated on 20/Oct/21
⇒1+((1−cos 2x)/2)=((3sin 2x)/2) ⇒3−cos 2x=3sin 2x ⇒3−cos 2x=3(√(1−cos^2 (2x))) ⇒9+cos^2 (2x)−6cos (2x)=9−9cos^2 (2x) ⇒10cos^2 (2x)−6cos (2x)=0 ⇒cos 2x(5cos (2x)−3)=0 { ((cos 2x=0⇒x=±(π/4)+nπ)),((cos 2x=(3/5)⇒tan (2x)=(4/3))) :} ⇒((2tan x)/(1−tan^2 x))=(4/3) ⇒6tan x=4−4tan^2 x ⇒2tan^2 x+3tan x−2=0 ⇒(2tan x−1)(tan x+2)=0 Σ tan (x_i )= −(3/2)
$$\Rightarrow\mathrm{1}+\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{2}}=\frac{\mathrm{3sin}\:\mathrm{2}{x}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{3}−\mathrm{cos}\:\mathrm{2}{x}=\mathrm{3sin}\:\mathrm{2}{x} \\ $$$$\Rightarrow\mathrm{3}−\mathrm{cos}\:\mathrm{2}{x}=\mathrm{3}\sqrt{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{2}{x}\right)} \\ $$$$\Rightarrow\mathrm{9}+\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{2}{x}\right)−\mathrm{6cos}\:\left(\mathrm{2}{x}\right)=\mathrm{9}−\mathrm{9cos}\:^{\mathrm{2}} \left(\mathrm{2}{x}\right) \\ $$$$\Rightarrow\mathrm{10cos}\:^{\mathrm{2}} \left(\mathrm{2}{x}\right)−\mathrm{6cos}\:\left(\mathrm{2}{x}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{cos}\:\mathrm{2}{x}\left(\mathrm{5cos}\:\left(\mathrm{2}{x}\right)−\mathrm{3}\right)=\mathrm{0} \\ $$$$\:\begin{cases}{\mathrm{cos}\:\mathrm{2}{x}=\mathrm{0}\Rightarrow{x}=\pm\frac{\pi}{\mathrm{4}}+{n}\pi}\\{\mathrm{cos}\:\mathrm{2}{x}=\frac{\mathrm{3}}{\mathrm{5}}\Rightarrow\mathrm{tan}\:\left(\mathrm{2}{x}\right)=\frac{\mathrm{4}}{\mathrm{3}}}\end{cases} \\ $$$$\:\Rightarrow\frac{\mathrm{2tan}\:{x}}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {x}}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{6tan}\:{x}=\mathrm{4}−\mathrm{4tan}\:^{\mathrm{2}} {x} \\ $$$$\Rightarrow\mathrm{2tan}\:^{\mathrm{2}} {x}+\mathrm{3tan}\:{x}−\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{2tan}\:{x}−\mathrm{1}\right)\left(\mathrm{tan}\:{x}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\Sigma\:\mathrm{tan}\:\left({x}_{{i}} \right)=\:−\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Commented by mr W last updated on 19/Oct/21
tan x=(1/2) or 1 ⇒Σtan x=(3/2)
$$\mathrm{tan}\:{x}=\frac{\mathrm{1}}{\mathrm{2}}\:{or}\:\mathrm{1} \\ $$$$\Rightarrow\Sigma\mathrm{tan}\:{x}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Answered by mr W last updated on 19/Oct/21
1+(1/2)(1−cos 2x)=(3/2) sin 2x 3 sin 2x+cos 2x=3 let t=tan x 3×((2t)/(1+t^2 ))+((1−t^2 )/(1+t^2 ))=3 2t^2 −3t+1=0 Σt=Σtan x=(3/2)
$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}\right)=\frac{\mathrm{3}}{\mathrm{2}}\:\mathrm{sin}\:\mathrm{2}{x} \\ $$$$\mathrm{3}\:\mathrm{sin}\:\mathrm{2}{x}+\mathrm{cos}\:\mathrm{2}{x}=\mathrm{3} \\ $$$${let}\:{t}=\mathrm{tan}\:{x} \\ $$$$\mathrm{3}×\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }+\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }=\mathrm{3} \\ $$$$\mathrm{2}{t}^{\mathrm{2}} −\mathrm{3}{t}+\mathrm{1}=\mathrm{0} \\ $$$$\Sigma{t}=\Sigma\mathrm{tan}\:{x}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Answered by amin96 last updated on 19/Oct/21
(1/(cos^2 (x)))+((sin^2 (x))/(cos^2 (x)))=3((sin(x)cos(x))/(cos^2 (x))) 1+tg^2 (x)+t^2 (x)=3tg(x) 2tg^2 (x)−3tg(x)+1=0 2a^2 −3a+1=0 { ((a_1 =tg(x)_1 =(1/2))),((a_2 =tg(x)_2 =1)) :} a_1 +a_2 =1,5 I think a solution like this would be easier.
$$\frac{\mathrm{1}}{\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{x}}\right)}+\frac{\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{x}}\right)}{\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{x}}\right)}=\mathrm{3}\frac{\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{x}}\right)}{\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{x}}\right)} \\ $$$$\mathrm{1}+\boldsymbol{\mathrm{tg}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{x}}\right)+\boldsymbol{\mathrm{t}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{x}}\right)=\mathrm{3}\boldsymbol{\mathrm{tg}}\left(\boldsymbol{\mathrm{x}}\right) \\ $$$$\mathrm{2}\boldsymbol{\mathrm{tg}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{x}}\right)−\mathrm{3}\boldsymbol{\mathrm{tg}}\left(\boldsymbol{\mathrm{x}}\right)+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{2}\boldsymbol{\mathrm{a}}^{\mathrm{2}} −\mathrm{3}\boldsymbol{\mathrm{a}}+\mathrm{1}=\mathrm{0}\:\:\:\begin{cases}{\boldsymbol{\mathrm{a}}_{\mathrm{1}} =\boldsymbol{\mathrm{tg}}\left(\boldsymbol{\mathrm{x}}\right)_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}}\\{\boldsymbol{\mathrm{a}}_{\mathrm{2}} =\boldsymbol{\mathrm{tg}}\left(\boldsymbol{\mathrm{x}}\right)_{\mathrm{2}} =\mathrm{1}}\end{cases} \\ $$$$\boldsymbol{\mathrm{a}}_{\mathrm{1}} +\boldsymbol{\mathrm{a}}_{\mathrm{2}} =\mathrm{1},\mathrm{5}\:\:\: \\ $$I think a solution like this would be easier.
Commented by amin96 last updated on 19/Oct/21
just occurred to me)
$$ \\ $$just occurred to me)
Answered by lyubita last updated on 19/Oct/21
sin^2 x + cos^2 x + sin^2 x = 3 sin x cos x 2 sin^2 x − 3 sin x cos x + cos^2 x = 0 (2 sin x − cos x)(sin x − cos x) = 0 2 sin x = cos x sin x = cos x tan x = (1/2) tan x = 1
$$\mathrm{sin}^{\mathrm{2}} \:{x}\:+\:\mathrm{cos}^{\mathrm{2}} \:{x}\:+\:\mathrm{sin}^{\mathrm{2}} \:{x}\:=\:\mathrm{3}\:\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}\: \\ $$$$\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:{x}\:−\:\mathrm{3}\:\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}\:+\:\mathrm{cos}^{\mathrm{2}} \:{x}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{2}\:\mathrm{sin}\:{x}\:−\:\mathrm{cos}\:{x}\right)\left(\mathrm{sin}\:{x}\:−\:\mathrm{cos}\:{x}\right)\:=\:\mathrm{0} \\ $$$$\:\:\:\mathrm{2}\:\mathrm{sin}\:{x}\:=\:\mathrm{cos}\:{x}\:\:\:\:\:\:\:\:\mathrm{sin}\:{x}\:=\:\mathrm{cos}\:{x} \\ $$$$\:\:\:\:\:\:\:\mathrm{tan}\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{tan}\:{x}\:=\:\mathrm{1} \\ $$$$\: \\ $$

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