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Question Number 105650 by bemath last updated on 30/Jul/20

cos^(−1) (2x)+ cos^(−1) (x)=(π/6)  find x

cos1(2x)+cos1(x)=π6findx

Answered by bramlex last updated on 30/Jul/20

cos^(−1) (2x)=(π/6)−cos^(−1) (x)  cos (cos^(−1) (2x))= cos ((π/6)−cos^(−1) (x))  2x = ((√3)/2) x+(1/2)sin (cos^(−1) (x))  4x=(√3) x+sin (cos^(−1) (x))  x(4−(√3))=sin (cos^(−1) (x))  x(4−(√3)) = (√(1−cos^2 (cos^(−1) (x))))  x(4−(√3)) = (√(1−x^2 ))  x^2 (4−(√3))^2 =1−x^2   x^2 (19−8(√3)+1) = 1  x^2 = (1/(20−8(√3))) = (1/(4(5−2(√3))))  x = ±(1/(2(√(5−2(√3))))) .▲

cos1(2x)=π6cos1(x)cos(cos1(2x))=cos(π6cos1(x))2x=32x+12sin(cos1(x))4x=3x+sin(cos1(x))x(43)=sin(cos1(x))x(43)=1cos2(cos1(x))x(43)=1x2x2(43)2=1x2x2(1983+1)=1x2=12083=14(523)x=±12523.

Answered by mathmax by abdo last updated on 31/Jul/20

⇒cos(arcos(2x)+cosx) =((√3)/2) ⇒  2x .x −sin(arcos(2x))sin(cosx) =((√3)/2) ⇒  2x^2 −(√(1−(2x)^2 ))×(√(1−x^2 ))=((√3)/2) ⇒  (2x^2 −((√3)/2)) =(√(1−4x^2 ))×(√(1−x^2 )) ⇒  (2x^2 −((√3)/2))^2  =(1−4x^2 )(1−x^2 ) ⇒  4x^4 −2(√3)x^2  +(3/4) =1−x^2 −4x^2 +4x^4  ⇒  −2(√3)x^2  +(3/4) =1−5x^2  ⇒−8(√3)x^2 +3 =4−20x^2  ⇒  (20−8(√3))x^2  =−3 ⇒(8(√3)−20)x^2  =3 ⇒x^2  =(3/(8(√3)−20)) ⇒  x =+^− (√(3/(8(√3)−20)))

cos(arcos(2x)+cosx)=322x.xsin(arcos(2x))sin(cosx)=322x21(2x)2×1x2=32(2x232)=14x2×1x2(2x232)2=(14x2)(1x2)4x423x2+34=1x24x2+4x423x2+34=15x283x2+3=420x2(2083)x2=3(8320)x2=3x2=38320x=+38320

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