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Question Number 59389 by hovea cw last updated on 09/May/19

If   0≤ x ≤ π  and  81^(sin^2 x) + 81^(cos^2 x) =30,  then  x is equal to

If0xπand81sin2x+81cos2x=30,thenxisequalto

Answered by ajfour last updated on 09/May/19

   t+((81)/t)=30  ⇒  t^2 −30t+81=0    ⇒    t=3, 27     81^(sin^2 x) =81^(1/4) , 81^(3/4)      ⇒   x=(π/6), (π/3) , ((2π)/3), ((5π)/6) .

t+81t=30t230t+81=0t=3,2781sin2x=8114,8134x=π6,π3,2π3,5π6.

Answered by tanmay last updated on 09/May/19

(81)^(sin^2 x) +(81)^(1−sin^2 x) =30  (81)^(sin^2 x) +((81)/((81)^(sin^2 x) ))=30  (9)^(2sin^2 x) +(9^2 /((9)^(2sin^2 x) ))=30  (9^(sin^2 x) −(9/9^(sin^2 x) ))^2 +2×9^(sin^2 x) ×(9/9^(sin^2 x) )=30  (9^(sin^2 x) −(9/9^(sin^2 x) ))^2 =(2(√3) )^2   (k−(9/k))^2 −(2(√3) )^2 =0  k−(9/k)+2(√3) =0  k^2 +2(√3) k−9=0  k^2 +(3(√3) −(√3) )k −3×(√3) ×(√3) =0  k(k+3(√3) )−(√3) (k+3(√3) )=0  (k+3(√3) )(k−(√3) )=0  k+3(√3) ≠0  k=(√3)   9^(sin^2 x) =3^(1/2)   3^(2sin^2 x) =3^(1/2)   sin^2 x=(1/4)  sinx=(1/2)=sin((π/6))→x=(π/6)(first quadrant)  and (π−(π/6))i,e ((5π)/6) in second quadrant  sinx≠−(1/2)  when π≥x≥0  again  k−(9/k)−2(√3) =0  k^2 −9−2(√3) k=0  k^2 −2(√3) k+3−12=0  (k−(√3) )^2 −(2(√3) )^2 =0  (k−(√3) +2(√3) )(k−(√3) −2(√3) )=0  k+(√3) ≠0  k=3(√3)   9^(sin^2 x) =3^(3/2)   3^(2sin^2 x) =3^(3/2)   sin^2 x=(3/4)  sinx=((√3)/2)→x=(π/3)and (π−(π/3))i,e ((2π)/3)  hence x=(π/6),(π/3),((2π)/3),((5π)/6)  since  π≥x≥0   so sinx≠ negetive  so sinx≠((−(√3))/2)

(81)sin2x+(81)1sin2x=30(81)sin2x+81(81)sin2x=30(9)2sin2x+92(9)2sin2x=30(9sin2x99sin2x)2+2×9sin2x×99sin2x=30(9sin2x99sin2x)2=(23)2(k9k)2(23)2=0k9k+23=0k2+23k9=0k2+(333)k3×3×3=0k(k+33)3(k+33)=0(k+33)(k3)=0k+330k=39sin2x=31232sin2x=312sin2x=14sinx=12=sin(π6)x=π6(firstquadrant)and(ππ6)i,e5π6insecondquadrantsinx12whenπx0againk9k23=0k2923k=0k223k+312=0(k3)2(23)2=0(k3+23)(k323)=0k+30k=339sin2x=33232sin2x=332sin2x=34sinx=32x=π3and(ππ3)i,e2π3hencex=π6,π3,2π3,5π6sinceπx0sosinxnegetivesosinx32

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