Question-190508

Question Number 190508 by 073 last updated on 04/Apr/23

Answered by Frix last updated on 04/Apr/23

t=tan x ((t+1)/( (√(t^2 +1))))=(1/2) 4(t+1)^2 =t^2 +1 t^2 +((8t)/3)+1=0 tan x =t=−(4/3)+((√7)/3) [testing ⇒ only 1 solution]

$${t}=\mathrm{tan}\:{x} \\ $$$$\frac{{t}+\mathrm{1}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{4}\left({t}+\mathrm{1}\right)^{\mathrm{2}} ={t}^{\mathrm{2}} +\mathrm{1} \\ $$$${t}^{\mathrm{2}} +\frac{\mathrm{8}{t}}{\mathrm{3}}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{tan}\:{x}\:={t}=−\frac{\mathrm{4}}{\mathrm{3}}+\frac{\sqrt{\mathrm{7}}}{\mathrm{3}}\:\left[\mathrm{testing}\:\Rightarrow\:\mathrm{only}\:\mathrm{1}\:\mathrm{solution}\right] \\ $$

Commented by 073 last updated on 04/Apr/23

$$\mathrm{nice}\:\mathrm{solution} \\ $$

Answered by mehdee42 last updated on 04/Apr/23

solution 1 if tan(x/2)=u⇒((2u)/(1+u^2 ))+((1−u^2 )/(1+u^2 ))=(1/2) 3u^2 −4u−1=0⇒u=((2±(√7))/3) tanx=((2u)/(1−u^2 )) .... solution 2 1+2sinxcosx=(1/4)⇒sin2x=−(3/4) ⇒cos2x=±((√7)/4) tan^2 x=((1+cos2x)/(1−cos2x))⇒tanx=±....

$${solution}\:\:\mathrm{1} \\ $$$$\:{if}\:\:{tan}\frac{{x}}{\mathrm{2}}={u}\Rightarrow\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }+\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{3}{u}^{\mathrm{2}} −\mathrm{4}{u}−\mathrm{1}=\mathrm{0}\Rightarrow{u}=\frac{\mathrm{2}\pm\sqrt{\mathrm{7}}}{\mathrm{3}} \\ $$$${tanx}=\frac{\mathrm{2}{u}}{\mathrm{1}−{u}^{\mathrm{2}} }\:\:\:\:\:…. \\ $$$${solution}\:\:\mathrm{2} \\ $$$$\mathrm{1}+\mathrm{2}{sinxcosx}=\frac{\mathrm{1}}{\mathrm{4}}\Rightarrow{sin}\mathrm{2}{x}=−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow{cos}\mathrm{2}{x}=\pm\frac{\sqrt{\mathrm{7}}}{\mathrm{4}} \\ $$$${tan}^{\mathrm{2}} {x}=\frac{\mathrm{1}+{cos}\mathrm{2}{x}}{\mathrm{1}−{cos}\mathrm{2}{x}}\Rightarrow{tanx}=\pm…. \\ $$$$ \\ $$$$ \\ $$

Answered by cortano12 last updated on 04/Apr/23

Given sin x+cos x=(1/2) (√2) ((1/( (√2))) sin x+(1/2)(√2) cos x)=(1/2) sin (x+45°)= (1/(2(√2))) tan (x+45°)=(1/( (√7))) ⇒((1+tan x)/(1−tan x)) = (1/( (√7))) ⇒(√7) +(√7) tan x = 1−tan x ⇒tan x = ((1−(√7))/(1+(√7))) = ((8−2(√7))/(−6)) ⇒tan x = (((√7)−4)/3)

$$\:\mathrm{Given}\:\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\sqrt{\mathrm{2}}\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\mathrm{sin}\:\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}}\:\mathrm{cos}\:\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\mathrm{sin}\:\left(\mathrm{x}+\mathrm{45}°\right)=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\: \\ $$$$\:\mathrm{tan}\:\left(\mathrm{x}+\mathrm{45}°\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}}} \\ $$$$\:\Rightarrow\frac{\mathrm{1}+\mathrm{tan}\:\mathrm{x}}{\mathrm{1}−\mathrm{tan}\:\mathrm{x}}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}}} \\ $$$$\Rightarrow\sqrt{\mathrm{7}}\:+\sqrt{\mathrm{7}}\:\mathrm{tan}\:\mathrm{x}\:=\:\mathrm{1}−\mathrm{tan}\:\mathrm{x} \\ $$$$\Rightarrow\mathrm{tan}\:\mathrm{x}\:=\:\frac{\mathrm{1}−\sqrt{\mathrm{7}}}{\mathrm{1}+\sqrt{\mathrm{7}}}\:=\:\frac{\mathrm{8}−\mathrm{2}\sqrt{\mathrm{7}}}{−\mathrm{6}} \\ $$$$\Rightarrow\mathrm{tan}\:\mathrm{x}\:=\:\frac{\sqrt{\mathrm{7}}−\mathrm{4}}{\mathrm{3}} \\ $$

Commented by mehdee42 last updated on 04/Apr/23

$${bravo} \\ $$

Commented by Spillover last updated on 05/Apr/23

$$\mathrm{great} \\ $$

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