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Question Number 171719 by cortano1 last updated on 20/Jun/22
(1/(1+sin^2 x)) + (1/(1+cos^2 x)) = ((48)/(35))
$$\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{sin}\:^{\mathrm{2}} {x}}\:+\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} {x}}\:=\:\frac{\mathrm{48}}{\mathrm{35}} \\ $$
Commented by infinityaction last updated on 20/Jun/22
sin x = ±(1/2),±((√3)/2)
$$\mathrm{sin}\:{x}\:=\:\pm\frac{\mathrm{1}}{\mathrm{2}},\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$
Commented by infinityaction last updated on 20/Jun/22
(3/((1+sin^2 x)(1+cos^2 x))) = ((48)/(35)) (1+sin^2 x)(2−sin^2 x) = ((35)/(16)) sin^2 x = z (1+z)(2−z) = ((35)/(16)) 16z^2 −16z+3 = 0 z = ((16±(√(256−192)))/(32)) z = (1/4) , (3/4) so z = sin^2 x = (1/4) ⇒ sinx = ±(1/2) and sin x = ±((√3)/2)
$$\:\:\:\:\:\:\frac{\mathrm{3}}{\left(\mathrm{1}+\mathrm{sin}\:^{\mathrm{2}} {x}\right)\left(\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} {x}\right)}\:=\:\frac{\mathrm{48}}{\mathrm{35}} \\ $$$$\:\:\:\:\:\left(\mathrm{1}+\mathrm{sin}\:^{\mathrm{2}} {x}\right)\left(\mathrm{2}−\mathrm{sin}\:^{\mathrm{2}} {x}\right)\:=\:\frac{\mathrm{35}}{\mathrm{16}} \\ $$$$\:\:\:\:\:\:\mathrm{sin}\:^{\mathrm{2}} {x}\:=\:{z} \\ $$$$\:\:\:\:\:\left(\mathrm{1}+{z}\right)\left(\mathrm{2}−{z}\right)\:=\:\frac{\mathrm{35}}{\mathrm{16}} \\ $$$$\:\:\:\:\:\:\mathrm{16}{z}^{\mathrm{2}} −\mathrm{16}{z}+\mathrm{3}\:=\:\mathrm{0} \\ $$$$\:\:\:\:\:{z}\:=\:\frac{\mathrm{16}\pm\sqrt{\mathrm{256}−\mathrm{192}}}{\mathrm{32}} \\ $$$$\:\:\:\:\:{z}\:=\:\:\frac{\mathrm{1}}{\mathrm{4}}\:,\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\:\:\:{so} \\ $$$$\:\:\:{z}\:=\:\mathrm{sin}\:^{\mathrm{2}} {x}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\:\:\Rightarrow\:\:\mathrm{sin}{x}\:\:=\:\:\:\pm\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:{and}\:\:\:\:\mathrm{sin}\:{x}\:\:=\:\:\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$
Answered by floor(10²Eta[1]) last updated on 20/Jun/22
(((1+cos^2 x)+(1+sin^2 x))/((1+sin^2 x)(1+cos^2 x)))=((48)/(35)) (3/(1+sin^2 x+cos^2 x+sin^2 xcos^2 x))=((48)/(35)) 35=16(2+(sinxcosx)^2 ) ((35)/(16))=2+((sin^2 (2x))/4) (3/4)=sin^2 (2x)⇒sin(2x)=±((√3)/2) 2x∈{(π/3),((2π)/3),((4π)/3),((5π)/3)}, 2x∈[0,2π] ⇒x∈{(π/6),(π/3),((2π)/3),((5π)/6)}, x∈[0,π]
$$\frac{\left(\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \mathrm{x}\right)+\left(\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \mathrm{x}\right)\left(\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \mathrm{x}\right)}=\frac{\mathrm{48}}{\mathrm{35}} \\ $$$$\frac{\mathrm{3}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \mathrm{x}+\mathrm{cos}^{\mathrm{2}} \mathrm{x}+\mathrm{sin}^{\mathrm{2}} \mathrm{xcos}^{\mathrm{2}} \mathrm{x}}=\frac{\mathrm{48}}{\mathrm{35}} \\ $$$$\mathrm{35}=\mathrm{16}\left(\mathrm{2}+\left(\mathrm{sinxcosx}\right)^{\mathrm{2}} \right) \\ $$$$\frac{\mathrm{35}}{\mathrm{16}}=\mathrm{2}+\frac{\mathrm{sin}^{\mathrm{2}} \left(\mathrm{2x}\right)}{\mathrm{4}} \\ $$$$\frac{\mathrm{3}}{\mathrm{4}}=\mathrm{sin}^{\mathrm{2}} \left(\mathrm{2x}\right)\Rightarrow\mathrm{sin}\left(\mathrm{2x}\right)=\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{2x}\in\left\{\frac{\pi}{\mathrm{3}},\frac{\mathrm{2}\pi}{\mathrm{3}},\frac{\mathrm{4}\pi}{\mathrm{3}},\frac{\mathrm{5}\pi}{\mathrm{3}}\right\},\:\mathrm{2x}\in\left[\mathrm{0},\mathrm{2}\pi\right] \\ $$$$\Rightarrow\mathrm{x}\in\left\{\frac{\pi}{\mathrm{6}},\frac{\pi}{\mathrm{3}},\frac{\mathrm{2}\pi}{\mathrm{3}},\frac{\mathrm{5}\pi}{\mathrm{6}}\right\},\:\mathrm{x}\in\left[\mathrm{0},\pi\right] \\ $$

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