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sin-10-x-cos-10-x-61-256-




Question Number 140382 by john_santu last updated on 07/May/21
sin^(10) x + cos^(10) x = ((61)/(256))
$$\mathrm{sin}\:^{\mathrm{10}} {x}\:+\:\mathrm{cos}\:^{\mathrm{10}} {x}\:=\:\frac{\mathrm{61}}{\mathrm{256}} \\ $$
Answered by MJS_new last updated on 07/May/21
sin x =s∧cos x =(√(1−s^2 ))  5s^8 −10s^6 +20s^4 −5s^2 +1=((61)/(256))  s^8 −2s^6 +2s^4 −s^2 +((39)/(256))=0  ⇒  sin x =±(1/2) ∨ sin x =±((√3)/2)  within 0≤x<2π we get  x∈{(π/6), (π/3), ((2π)/3), ((5π)/6), ((7π)/6), ((4π)/3), ((5π)/3), ((11π)/6)}
$$\mathrm{sin}\:{x}\:={s}\wedge\mathrm{cos}\:{x}\:=\sqrt{\mathrm{1}−{s}^{\mathrm{2}} } \\ $$$$\mathrm{5}{s}^{\mathrm{8}} −\mathrm{10}{s}^{\mathrm{6}} +\mathrm{20}{s}^{\mathrm{4}} −\mathrm{5}{s}^{\mathrm{2}} +\mathrm{1}=\frac{\mathrm{61}}{\mathrm{256}} \\ $$$${s}^{\mathrm{8}} −\mathrm{2}{s}^{\mathrm{6}} +\mathrm{2}{s}^{\mathrm{4}} −{s}^{\mathrm{2}} +\frac{\mathrm{39}}{\mathrm{256}}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\mathrm{sin}\:{x}\:=\pm\frac{\mathrm{1}}{\mathrm{2}}\:\vee\:\mathrm{sin}\:{x}\:=\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{within}\:\mathrm{0}\leqslant{x}<\mathrm{2}\pi\:\mathrm{we}\:\mathrm{get} \\ $$$${x}\in\left\{\frac{\pi}{\mathrm{6}},\:\frac{\pi}{\mathrm{3}},\:\frac{\mathrm{2}\pi}{\mathrm{3}},\:\frac{\mathrm{5}\pi}{\mathrm{6}},\:\frac{\mathrm{7}\pi}{\mathrm{6}},\:\frac{\mathrm{4}\pi}{\mathrm{3}},\:\frac{\mathrm{5}\pi}{\mathrm{3}},\:\frac{\mathrm{11}\pi}{\mathrm{6}}\right\} \\ $$
Commented by benjo_mathlover last updated on 07/May/21
waw...master of trigonometry
$$\mathrm{waw}…\mathrm{master}\:\mathrm{of}\:\mathrm{trigonometry} \\ $$
Commented by greg_ed last updated on 07/May/21
ok !
$$\mathrm{ok}\:! \\ $$

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