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Question Number 173787 by mnjuly1970 last updated on 18/Jul/22

4sin^( 2) (x)− 4sin(x)=cos^( 2) (x)−2cos(x) , 0<x<(π/2) . Find sin(x)=?

$$\:\mathrm{4}{sin}^{\:\mathrm{2}} \left({x}\right)−\:\mathrm{4}{sin}\left({x}\right)={cos}^{\:\mathrm{2}} \left({x}\right)−\mathrm{2}{cos}\left({x}\right) \\ $$$$\:\:\:,\:\mathrm{0}<{x}<\frac{\pi}{\mathrm{2}}\:.\:\:\:{Find}\:\:\:\:\:\:\:{sin}\left({x}\right)=? \\ $$

Commented by infinityaction last updated on 18/Jul/22

(1/( (√5)))

$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$

Commented by som(math1967) last updated on 18/Jul/22

(3/5) also

$$\frac{\mathrm{3}}{\mathrm{5}}\:{also} \\ $$

Commented by infinityaction last updated on 18/Jul/22

4sin^2 x−4sinx+1 = cos^2 x−2cos x+1 (1−2sinx)^2 = (1−cosx)^2 (1−2sinx)^2 − (1−cosx)^2 = 0 (cosx−2sinx)(2−2sinx−cos x)= 0 if cosx−2sinx = 0 ⇒ tanx = (1/2) sinx = (1/( (√5))) 2−2sinx = (√(1−sin^2 x )) 5sin^2 x −8sinx +3 = 0 (sinx−1)(5sinx−3)=0 sinx = 1 (rejected) sinx = (3/5)

$$\:\:\:\mathrm{4sin}^{\mathrm{2}} {x}−\mathrm{4sin}{x}+\mathrm{1}\:=\:\mathrm{cos}^{\mathrm{2}} {x}−\mathrm{2cos}\:{x}+\mathrm{1} \\ $$$$\:\:\:\left(\mathrm{1}−\mathrm{2sin}{x}\right)^{\mathrm{2}} \:\:=\:\:\left(\mathrm{1}−\mathrm{cos}{x}\right)^{\mathrm{2}} \:\:\: \\ $$$$\:\:\:\left(\mathrm{1}−\mathrm{2sin}{x}\right)^{\mathrm{2}} \:−\:\left(\mathrm{1}−\mathrm{cos}{x}\right)^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\:\:\left(\mathrm{cos}{x}−\mathrm{2sin}{x}\right)\left(\mathrm{2}−\mathrm{2sin}{x}−\mathrm{cos}\:{x}\right)=\:\mathrm{0}\:\: \\ $$$$\:\:\:{if}\:\:\:\mathrm{cos}{x}−\mathrm{2sin}{x}\:=\:\mathrm{0}\:\Rightarrow\:\mathrm{tan}{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\: \\ $$$$\:\:\:\:\:\mathrm{sin}{x}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:\:\: \\ $$$$\:\:\:\:\mathrm{2}−\mathrm{2sin}{x}\:\:=\:\:\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} {x}\:}\: \\ $$$$\:\:\:\:\mathrm{5sin}^{\mathrm{2}} {x}\:−\mathrm{8sin}{x}\:+\mathrm{3}\:=\:\mathrm{0} \\ $$$$\:\:\:\left(\mathrm{sin}{x}−\mathrm{1}\right)\left(\mathrm{5sin}{x}−\mathrm{3}\right)=\mathrm{0} \\ $$$$\:\:\:\:\mathrm{sin}{x}\:=\:\mathrm{1}\:\:\left({rejected}\right) \\ $$$$\:\:\:\:\mathrm{sin}{x}\:=\:\frac{\mathrm{3}}{\mathrm{5}}\:\: \\ $$

Commented by mnjuly1970 last updated on 18/Jul/22

thank you so much sir..

$$\:\:{thank}\:{you}\:{so}\:{much}\:{sir}.. \\ $$

Commented by MJS_new last updated on 18/Jul/22

why do you reject 1?

$$\mathrm{why}\:\mathrm{do}\:\mathrm{you}\:\mathrm{reject}\:\mathrm{1}? \\ $$

Commented by infinityaction last updated on 18/Jul/22

given in the question 0<x<(π/2)

$${given}\:{in}\:{the}\:{question} \\ $$$$\mathrm{0}<{x}<\frac{\pi}{\mathrm{2}} \\ $$

Commented by Tawa11 last updated on 18/Jul/22

Great sirs

$$\mathrm{Great}\:\mathrm{sirs} \\ $$

Commented by MJS_new last updated on 18/Jul/22

ok didn′t notice...

$$\mathrm{ok}\:\mathrm{didn}'\mathrm{t}\:\mathrm{notice}... \\ $$

Commented by infinityaction last updated on 18/Jul/22

no problem sir your algebra is very good i am student of undergruate so can you suggest me an algebra book ??

$$\:\:\:{no}\:{problem}\:{sir}\:{your}\:{algebra}\: \\ $$$$\:\:\:{is}\:{very}\:{good} \\ $$$$\:\:\:\:{i}\:{am}\:{student}\:{of}\:{undergruate} \\ $$$$\:\:\:\:{so}\:{can}\:{you}\:{suggest}\:{me}\:{an} \\ $$$$\:\:\:\:{algebra}\:{book}\:??\:\:\: \\ $$

Commented by MJS_new last updated on 18/Jul/22

sorry I can′t. I studied math more than 20 years ago but I became a musician. math is only my recreational sport now...

$$\mathrm{sorry}\:\mathrm{I}\:\mathrm{can}'\mathrm{t}.\:\mathrm{I}\:\mathrm{studied}\:\mathrm{math}\:\mathrm{more}\:\mathrm{than}\:\mathrm{20} \\ $$$$\mathrm{years}\:\mathrm{ago}\:\mathrm{but}\:\mathrm{I}\:\mathrm{became}\:\mathrm{a}\:\mathrm{musician}.\:\mathrm{math}\:\mathrm{is} \\ $$$$\mathrm{only}\:\mathrm{my}\:\mathrm{recreational}\:\mathrm{sport}\:\mathrm{now}... \\ $$

Commented by infinityaction last updated on 19/Jul/22

ok sir thanks

$${ok}\:{sir}\:{thanks} \\ $$

Answered by MJS_new last updated on 18/Jul/22

x=2arctan t ((−8t^3 +16t^2 −8t)/((t^2 +1)^2 ))=((3t^4 −2t^2 −1)/((t^2 +1)^2 )) −8t(t−1)^2 =(t−1)(t+1)(3t^2 +1) t_1 =1 3t^3 +11t^2 −7t+1=0 (3t−1)(t+2−(√5))(t+2+(√5))=0 t_2 =(1/3) t_3 =−2+(√5) t_4 =−2−(√5) sin x =sin 2arctan t =((2t)/(t^2 +1)) sin x ∈{−((√5)/5), ((√5)/5), (3/5), 1}

$${x}=\mathrm{2arctan}\:{t} \\ $$$$\frac{−\mathrm{8}{t}^{\mathrm{3}} +\mathrm{16}{t}^{\mathrm{2}} −\mathrm{8}{t}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{3}{t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{2}} −\mathrm{1}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$−\mathrm{8}{t}\left({t}−\mathrm{1}\right)^{\mathrm{2}} =\left({t}−\mathrm{1}\right)\left({t}+\mathrm{1}\right)\left(\mathrm{3}{t}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$${t}_{\mathrm{1}} =\mathrm{1} \\ $$$$\mathrm{3}{t}^{\mathrm{3}} +\mathrm{11}{t}^{\mathrm{2}} −\mathrm{7}{t}+\mathrm{1}=\mathrm{0} \\ $$$$\left(\mathrm{3}{t}−\mathrm{1}\right)\left({t}+\mathrm{2}−\sqrt{\mathrm{5}}\right)\left({t}+\mathrm{2}+\sqrt{\mathrm{5}}\right)=\mathrm{0} \\ $$$${t}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${t}_{\mathrm{3}} =−\mathrm{2}+\sqrt{\mathrm{5}} \\ $$$${t}_{\mathrm{4}} =−\mathrm{2}−\sqrt{\mathrm{5}} \\ $$$$\mathrm{sin}\:{x}\:=\mathrm{sin}\:\mathrm{2arctan}\:{t}\:=\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{sin}\:{x}\:\in\left\{−\frac{\sqrt{\mathrm{5}}}{\mathrm{5}},\:\frac{\sqrt{\mathrm{5}}}{\mathrm{5}},\:\frac{\mathrm{3}}{\mathrm{5}},\:\mathrm{1}\right\} \\ $$

Answered by a.lgnaoui last updated on 18/Jul/22

4sin^2 (x)−4sin (x)=1−sin^2 (x)−2(√(1−sin^2 (x) )) posins sin(x)=z 5z^2 −4z−1 =−2(√(1−z^2 )) [z−1 ]^2 [5z +1]^2 =4(1−z^2 ) (1−z)(5z+1)^2 =4(1+z) (1−z)(25z^2 +10z+1)=4z+4 25z^2 +10z+1−25z^3 −10z^2 −z−4z−4=0 −25z^3 +15z^2 −5z−3=0 25z^3 −15z^2 +5z+3=0 z^3 −(3/5)z^2 +(z/5)+(3/(25))=0 Z=z+1/5 Z^3 =z^3 +(3/5)z^2 +(3/(25))z+(1/(125)) −(3/5)Z^2 =((−3)/5)z^2 +−(6/(25))z+((−3)/(125)) Z=z+(1/5) Z^3 +((2/(25)))Z+((22)/(125))=0 p=(2/(25)) q=((22)/(125)) p^3 =(4×8/5^6 )/27+(4×11^2 )/ Δ=484/5^6 +((32)/(27×5^6 ))=((484×27+32)/(3^3 ×5^6 ))=((13068+32)/)=((13100)/(421875))=0,031⇒(√Δ)=0,176 q=((22)/(125))=0,176⇒(q/2)=0,088 =−q/2±[^3 (√(((−q)/2)±(4p^3 +27q^2 )/27)))] /2 (√(0,176−0,088)) Z=0,088+[^3 (√(0,088+0,176))]/2 Z= 0,088 +[^3 (√(−0,088+0,176))]/2 Z= 0,64 et Z=−0,56 z=Z−(1/5) ⇒ sin (x)=0,44 sin (x)=0,36

$$\mathrm{4sin}\:^{\mathrm{2}} \left(\mathrm{x}\right)−\mathrm{4sin}\:\left(\mathrm{x}\right)=\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)−\mathrm{2}\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)\:} \\ $$$${posins}\:\:\:\mathrm{sin}\left({x}\right)={z}\: \\ $$$$\mathrm{5}{z}^{\mathrm{2}} \:−\mathrm{4}{z}−\mathrm{1}\:=−\mathrm{2}\sqrt{\mathrm{1}−{z}^{\mathrm{2}} \:} \\ $$$$\left[{z}−\mathrm{1}\:\right]^{\mathrm{2}} \left[\mathrm{5}{z}\:+\mathrm{1}\right]^{\mathrm{2}} =\mathrm{4}\left(\mathrm{1}−{z}^{\mathrm{2}} \right)\: \\ $$$$\left(\mathrm{1}−{z}\right)\left(\mathrm{5}{z}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{4}\left(\mathrm{1}+{z}\right) \\ $$$$\left(\mathrm{1}−{z}\right)\left(\mathrm{25}{z}^{\mathrm{2}} +\mathrm{10}{z}+\mathrm{1}\right)=\mathrm{4}{z}+\mathrm{4} \\ $$$$\mathrm{25}{z}^{\mathrm{2}} +\mathrm{10}{z}+\mathrm{1}−\mathrm{25}{z}^{\mathrm{3}} −\mathrm{10}{z}^{\mathrm{2}} −{z}−\mathrm{4}{z}−\mathrm{4}=\mathrm{0} \\ $$$$−\mathrm{25}{z}^{\mathrm{3}} +\mathrm{15}{z}^{\mathrm{2}} −\mathrm{5}{z}−\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{25}{z}^{\mathrm{3}} −\mathrm{15}{z}^{\mathrm{2}} +\mathrm{5}{z}+\mathrm{3}=\mathrm{0} \\ $$$${z}^{\mathrm{3}} −\frac{\mathrm{3}}{\mathrm{5}}{z}^{\mathrm{2}} +\frac{{z}}{\mathrm{5}}+\frac{\mathrm{3}}{\mathrm{25}}=\mathrm{0} \\ $$$$\mathrm{Z}={z}+\mathrm{1}/\mathrm{5} \\ $$$$\mathrm{Z}^{\mathrm{3}} ={z}^{\mathrm{3}} +\frac{\mathrm{3}}{\mathrm{5}}{z}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{25}}{z}+\frac{\mathrm{1}}{\mathrm{125}} \\ $$$$−\frac{\mathrm{3}}{\mathrm{5}}\mathrm{Z}^{\mathrm{2}} =\frac{−\mathrm{3}}{\mathrm{5}}{z}^{\mathrm{2}} +−\frac{\mathrm{6}}{\mathrm{25}}{z}+\frac{−\mathrm{3}}{\mathrm{125}} \\ $$$$\mathrm{Z}={z}+\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\mathrm{Z}^{\mathrm{3}} +\left(\frac{\mathrm{2}}{\mathrm{25}}\right)\mathrm{Z}+\frac{\mathrm{22}}{\mathrm{125}}=\mathrm{0}\:\:\:\:\:\:{p}=\frac{\mathrm{2}}{\mathrm{25}}\:\:{q}=\frac{\mathrm{22}}{\mathrm{125}} \\ $$$${p}^{\mathrm{3}} =\left(\mathrm{4}×\mathrm{8}/\mathrm{5}^{\mathrm{6}} \right)/\mathrm{27}+\left(\mathrm{4}×\mathrm{11}^{\mathrm{2}} \right)/ \\ $$$$\Delta=\mathrm{484}/\mathrm{5}^{\mathrm{6}} +\frac{\mathrm{32}}{\mathrm{27}×\mathrm{5}^{\mathrm{6}} }=\frac{\mathrm{484}×\mathrm{27}+\mathrm{32}}{\mathrm{3}^{\mathrm{3}} ×\mathrm{5}^{\mathrm{6}} }=\frac{\mathrm{13068}+\mathrm{32}}{}=\frac{\mathrm{13100}}{\mathrm{421875}}=\mathrm{0},\mathrm{031}\Rightarrow\sqrt{\Delta}=\mathrm{0},\mathrm{176} \\ $$$${q}=\frac{\mathrm{22}}{\mathrm{125}}=\mathrm{0},\mathrm{176}\Rightarrow\frac{{q}}{\mathrm{2}}=\mathrm{0},\mathrm{088} \\ $$$$ \\ $$$$=−{q}/\mathrm{2}\pm\left[\:^{\mathrm{3}} \sqrt{\left.\frac{−{q}}{\mathrm{2}}\pm\left(\mathrm{4}{p}^{\mathrm{3}} +\mathrm{27}{q}^{\mathrm{2}} \right)/\mathrm{27}\right)}\right]\:\:/\mathrm{2}\:\:\:\:\:\:\sqrt{\mathrm{0},\mathrm{176}−\mathrm{0},\mathrm{088}} \\ $$$$\mathrm{Z}=\mathrm{0},\mathrm{088}+\left[^{\mathrm{3}} \sqrt{\mathrm{0},\mathrm{088}+\mathrm{0},\mathrm{176}}\right]/\mathrm{2} \\ $$$$\mathrm{Z}=\:\mathrm{0},\mathrm{088}\:+\left[\:^{\mathrm{3}} \sqrt{−\mathrm{0},\mathrm{088}+\mathrm{0},\mathrm{176}}\right]/\mathrm{2} \\ $$$$\:\mathrm{Z}=\:\:\mathrm{0},\mathrm{64}\:\:\:\:\:\:\:\:\:\:\:\:\:{et}\:\:\:\:\:\:\:\mathrm{Z}=−\mathrm{0},\mathrm{56} \\ $$$${z}=\mathrm{Z}−\frac{\mathrm{1}}{\mathrm{5}}\:\:\Rightarrow\:\:\mathrm{sin}\:\left({x}\right)=\mathrm{0},\mathrm{44} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{sin}\:\left({x}\right)=\mathrm{0},\mathrm{36}\:\:\:\:\:\:\:\: \\ $$

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