Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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 byinfinityaction last updated on 18/Jul/22

(1/( (√5)))

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

Commented bysom(math1967) last updated on 18/Jul/22

(3/5) also

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

Commented byinfinityaction 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 bymnjuly1970 last updated on 18/Jul/22

  thank you so much sir..

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

Commented byMJS_new last updated on 18/Jul/22

why do you reject 1?

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

Commented byinfinityaction 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 byTawa11 last updated on 18/Jul/22

Great sirs

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

Commented byMJS_new last updated on 18/Jul/22

ok didn′t notice...

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

Commented byinfinityaction 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 byMJS_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 byinfinityaction 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

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com