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If-0-x-pi-and-81-sin-2-x-81-cos-2-x-30-then-x-is-equal-to-

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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
$$\mathrm{If}\:\:\:\mathrm{0}\leqslant\:{x}\:\leqslant\:\pi\:\:\mathrm{and}\:\:\mathrm{81}^{\mathrm{sin}^{\mathrm{2}} {x}} +\:\mathrm{81}^{\mathrm{cos}^{\mathrm{2}} {x}} =\mathrm{30}, \\ $$$$\mathrm{then}\:\:{x}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$
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) .
$$\:\:\:\mathrm{t}+\frac{\mathrm{81}}{\mathrm{t}}=\mathrm{30} \\ $$$$\Rightarrow\:\:\mathrm{t}^{\mathrm{2}} −\mathrm{30t}+\mathrm{81}=\mathrm{0} \\ $$$$\:\:\Rightarrow\:\:\:\:\mathrm{t}=\mathrm{3},\:\mathrm{27} \\ $$$$\:\:\:\mathrm{81}^{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}} =\mathrm{81}^{\frac{\mathrm{1}}{\mathrm{4}}} ,\:\mathrm{81}^{\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$$\:\:\:\Rightarrow\:\:\:\mathrm{x}=\frac{\pi}{\mathrm{6}},\:\frac{\pi}{\mathrm{3}}\:,\:\frac{\mathrm{2}\pi}{\mathrm{3}},\:\frac{\mathrm{5}\pi}{\mathrm{6}}\:. \\ $$$$\:\:\:\: \\ $$
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)
$$\left(\mathrm{81}\right)^{{sin}^{\mathrm{2}} {x}} +\left(\mathrm{81}\right)^{\mathrm{1}−{sin}^{\mathrm{2}} {x}} =\mathrm{30} \\ $$$$\left(\mathrm{81}\right)^{{sin}^{\mathrm{2}} {x}} +\frac{\mathrm{81}}{\left(\mathrm{81}\right)^{{sin}^{\mathrm{2}} {x}} }=\mathrm{30} \\ $$$$\left(\mathrm{9}\right)^{\mathrm{2}{sin}^{\mathrm{2}} {x}} +\frac{\mathrm{9}^{\mathrm{2}} }{\left(\mathrm{9}\right)^{\mathrm{2}{sin}^{\mathrm{2}} {x}} }=\mathrm{30} \\ $$$$\left(\mathrm{9}^{{sin}^{\mathrm{2}} {x}} −\frac{\mathrm{9}}{\mathrm{9}^{{sin}^{\mathrm{2}} {x}} }\right)^{\mathrm{2}} +\mathrm{2}×\mathrm{9}^{{sin}^{\mathrm{2}} {x}} ×\frac{\mathrm{9}}{\mathrm{9}^{{sin}^{\mathrm{2}} {x}} }=\mathrm{30} \\ $$$$\left(\mathrm{9}^{{sin}^{\mathrm{2}} {x}} −\frac{\mathrm{9}}{\mathrm{9}^{{sin}^{\mathrm{2}} {x}} }\right)^{\mathrm{2}} =\left(\mathrm{2}\sqrt{\mathrm{3}}\:\right)^{\mathrm{2}} \\ $$$$\left({k}−\frac{\mathrm{9}}{{k}}\right)^{\mathrm{2}} −\left(\mathrm{2}\sqrt{\mathrm{3}}\:\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${k}−\frac{\mathrm{9}}{{k}}+\mathrm{2}\sqrt{\mathrm{3}}\:=\mathrm{0} \\ $$$${k}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{3}}\:{k}−\mathrm{9}=\mathrm{0} \\ $$$${k}^{\mathrm{2}} +\left(\mathrm{3}\sqrt{\mathrm{3}}\:−\sqrt{\mathrm{3}}\:\right){k}\:−\mathrm{3}×\sqrt{\mathrm{3}}\:×\sqrt{\mathrm{3}}\:=\mathrm{0} \\ $$$${k}\left({k}+\mathrm{3}\sqrt{\mathrm{3}}\:\right)−\sqrt{\mathrm{3}}\:\left({k}+\mathrm{3}\sqrt{\mathrm{3}}\:\right)=\mathrm{0} \\ $$$$\left({k}+\mathrm{3}\sqrt{\mathrm{3}}\:\right)\left({k}−\sqrt{\mathrm{3}}\:\right)=\mathrm{0} \\ $$$${k}+\mathrm{3}\sqrt{\mathrm{3}}\:\neq\mathrm{0} \\ $$$${k}=\sqrt{\mathrm{3}}\: \\ $$$$\mathrm{9}^{{sin}^{\mathrm{2}} {x}} =\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\mathrm{3}^{\mathrm{2}{sin}^{\mathrm{2}} {x}} =\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${sin}^{\mathrm{2}} {x}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${sinx}=\frac{\mathrm{1}}{\mathrm{2}}={sin}\left(\frac{\pi}{\mathrm{6}}\right)\rightarrow{x}=\frac{\pi}{\mathrm{6}}\left({first}\:{quadrant}\right) \\ $$$${and}\:\left(\pi−\frac{\pi}{\mathrm{6}}\right){i},{e}\:\frac{\mathrm{5}\pi}{\mathrm{6}}\:{in}\:{second}\:{quadrant} \\ $$$${sinx}\neq−\frac{\mathrm{1}}{\mathrm{2}}\:\:{when}\:\pi\geqslant{x}\geqslant\mathrm{0} \\ $$$${again} \\ $$$${k}−\frac{\mathrm{9}}{{k}}−\mathrm{2}\sqrt{\mathrm{3}}\:=\mathrm{0} \\ $$$${k}^{\mathrm{2}} −\mathrm{9}−\mathrm{2}\sqrt{\mathrm{3}}\:{k}=\mathrm{0} \\ $$$${k}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{3}}\:{k}+\mathrm{3}−\mathrm{12}=\mathrm{0} \\ $$$$\left({k}−\sqrt{\mathrm{3}}\:\right)^{\mathrm{2}} −\left(\mathrm{2}\sqrt{\mathrm{3}}\:\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({k}−\sqrt{\mathrm{3}}\:+\mathrm{2}\sqrt{\mathrm{3}}\:\right)\left({k}−\sqrt{\mathrm{3}}\:−\mathrm{2}\sqrt{\mathrm{3}}\:\right)=\mathrm{0} \\ $$$${k}+\sqrt{\mathrm{3}}\:\neq\mathrm{0} \\ $$$${k}=\mathrm{3}\sqrt{\mathrm{3}}\: \\ $$$$\mathrm{9}^{{sin}^{\mathrm{2}} {x}} =\mathrm{3}^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$\mathrm{3}^{\mathrm{2}{sin}^{\mathrm{2}} {x}} =\mathrm{3}^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$${sin}^{\mathrm{2}} {x}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${sinx}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\rightarrow{x}=\frac{\pi}{\mathrm{3}}{and}\:\left(\pi−\frac{\pi}{\mathrm{3}}\right){i},{e}\:\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$${hence}\:{x}=\frac{\pi}{\mathrm{6}},\frac{\pi}{\mathrm{3}},\frac{\mathrm{2}\pi}{\mathrm{3}},\frac{\mathrm{5}\pi}{\mathrm{6}} \\ $$$${since}\:\:\pi\geqslant{x}\geqslant\mathrm{0}\:\:\:{so}\:{sinx}\neq\:{negetive} \\ $$$${so}\:{sinx}\neq\frac{−\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$ \\ $$

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