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1-a-2-16-4-a-a-2-9-3-a-4-a-2-2-a-2-x-x-x-1-2-14x-x-1-12-0-3-log-4-2-2x-3-cos-x-6sin-2-x-x-where-5pi-2-x-4pi-4-2cos-2-x-

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Question Number 117938 by bemath last updated on 14/Oct/20
(1) ((∣a^2 −16∣)/(4−a)) − ((∣a^2 −9∣)/(3+a)) − ((∣4−a^2 ∣)/(2−a)) =? (2)(((∣x∣ +x )/(x−1)))^2 −((14x)/(x−1)) + 12 = 0 (3) log _4 (2^(2x) −(√3) cos x−6sin^2 x ) = x where ((5π)/2) ≤x≤4π (4) ((2cos^2 x−(√3) cos x)/(log _4 (sin x))) = 0 , where −3π ≤x≤−((3π)/2)
(1)∣a2−16∣4−a−∣a2−9∣3+a−∣4−a2∣2−a=?(2)(∣x∣+xx−1)2−14xx−1+12=0(3)log4(22x−3cosx−6sin2x)=xwhere5π2⩽x⩽4π(4)2cos2x−3cosxlog4(sinx)=0,where−3π⩽x⩽−3π2
Answered by john santu last updated on 14/Oct/20
(1) ((∣(a−4)(a+4)∣)/(4−a)) −((∣(a+3)(a−3)∣)/(3+a)) −((∣(a−2)(a+2)∣)/(2−a)) = for a<−4 ∨ a > 4 , gives (((a−4)(a+4))/(4−a)) − (((a+3)(a−3))/(a+3)) −(((a−2)(a+2))/(2−a)) = −a−4−a+3+a+2 = 1−a for −2<a<2 , gives (((4−a)(4+a))/(4−a))−(((3−a)(3+a))/(3+a))−(((2−a)(2+a))/(2−a)) = 4+a−3+a−2−a = −1+ a
(1)∣(a−4)(a+4)∣4−a−∣(a+3)(a−3)∣3+a−∣(a−2)(a+2)∣2−a=fora<−4∨a>4,gives(a−4)(a+4)4−a−(a+3)(a−3)a+3−(a−2)(a+2)2−a=−a−4−a+3+a+2=1−afor−2<a<2,gives(4−a)(4+a)4−a−(3−a)(3+a)3+a−(2−a)(2+a)2−a=4+a−3+a−2−a=−1+a
Answered by john santu last updated on 14/Oct/20
(2)(((∣x∣+x)/(x−1)))^2 −((14x)/(x−1)) +12 = 0 for x < 0 ⇒∣x∣ = −x ⇒−((14x)/(x−1)) = −12 ; 7x = 6x−6 ⇒ x = −6 . for x ≥ 0 ⇒∣x∣ = x ⇒((4x^2 )/((x−1)^2 ))−((14x(x−1))/((x−1)^2 )) +12 = 0; x≠1 2x^2 −7x^2 +7x+6(x^2 −2x+1)=0 x^2 −5x+6 = 0 ; (x−3)(x−2)=0 x = 3 ∧ x=2 . Thus solution set is { −6, 2, 3 }
(2)(∣x∣+xx−1)2−14xx−1+12=0forx<0⇒∣x∣=−x⇒−14xx−1=−12;7x=6x−6⇒x=−6.forx⩾0⇒∣x∣=x⇒4x2(x−1)2−14x(x−1)(x−1)2+12=0;x≠12x2−7x2+7x+6(x2−2x+1)=0x2−5x+6=0;(x−3)(x−2)=0x=3∧x=2.Thussolutionsetis{−6,2,3}
Answered by john santu last updated on 14/Oct/20
(3)log _4 (2^(2x) −(√3) cos x−6sin^2 x) = x ⇒ 2^(2x) −(√3) cos x−6sin^2 x = 4^x ⇒−6(1−cos^2 x)−(√3) cos x = 0 ⇒6 cos^2 x−(√(3 )) cos x−6 =0 ⇒(3cos x−2(√3))(2cos x+(√3))=0 ⇒cos x= −((√3)/2) ⇒x = ± ((5π)/6) + 2kπ
(3)log4(22x−3cosx−6sin2x)=x⇒22x−3cosx−6sin2x=4x⇒−6(1−cos2x)−3cosx=0⇒6cos2x−3cosx−6=0⇒(3cosx−23)(2cosx+3)=0⇒cosx=−32⇒x=±5π6+2kπ
Answered by Lordose last updated on 14/Oct/20
3. 4^x −(√3)cosx−6(1−cos^2 x)=4^x 6cos^2 x−(√3)cosx−6=0 (u−((√3)/(12)))^2 = ((147)/(144)) {u=cosx} u=(((√3)±(√(147)))/(144)) u= −0.8660 or u= 1.1547 ∴ x= cos^(−1) (−0.8660) or cos^(−1) (1.1547) x = 135° or x= complex x=((3π)/4) +2𝛑n {n∈N}
3.4x−3cosx−6(1−cos2x)=4x6cos2x−3cosx−6=0(u−312)2=147144{u=cosx}u=3±147144u=−0.8660oru=1.1547∴x=cos−1(−0.8660)orcos−1(1.1547)x=135°orx=complexx=3π4+2πn{n∈N}
Commented by bemath last updated on 14/Oct/20
cos^(−1) (−0.8660) = 150° not 135° sir
cos−1(−0.8660)=150°not135°sir
Commented by Lordose last updated on 14/Oct/20
Ohh Thanks
OhhThanks
Answered by john santu last updated on 14/Oct/20
(4)2cos^2 x−(√3) cos x = 0 ∧ sin x > 0 cos x(2cos x−(√3)) = 0 ∧ sin x >0 { ((cos x=0)),((cos x=((√3)/2))) :} ∧ −((5π)/2)<x<−((3π)/2)
(4)2cos2x−3cosx=0∧sinx>0cosx(2cosx−3)=0∧sinx>0{cosx=0cosx=32∧−5π2<x<−3π2

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