Họ nghiệm của phương trình 16(sin8x + cos8x) = 17cos22x là

     
If $$sin^8(x)+cos^8(x)=48/128,$$ then find the value of $x$? I tried this by De Moivre"s theorem:

$$(cos heta+isin heta)^n=cos(n heta)+isin( heta)=e^in heta$$

But could not proceed further please help.


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Hint:

Use the identity:$$(a-b)^2+(a+b)^2=2(a^2+b^2)$$

So:

$$frac22left(sin^8(x)+cos^8(x) ight)=frac12left((sin^4(x)-cos^4(x))^2+(sin^4(x)+cos^4(x))^2 ight)$$

You can repeat this for all the terms with the form $a^2+b^2$, for the other terms simplify using trigonometric identities.


*

If you define $s=sin^2(x)$ your equation becomes$$s^4+(1-s)^4=frac 48128\2s^4-4s^3+6s^2-4s+1=frac 48128\s^4-2s^3+3s^2-2s+frac 516=0$$for which Alpha finds the ugly real solutions$$s=frac 12left(1pmsqrtsqrt 11-3 ight)$$ & the complex solutions$$s=frac 12left(1pm isqrtsqrt 11+3 ight)$$but that doesn"t seem very enlightening to lớn me.


*

Assuming $xin Bbb R.$ For brevity let $c=cos x$ & $s=sin x.$ Let $p=c^2s^2.$ We have $$ c^8+s^8=frac 38iff$$ $$ 1=(c^2+s^2)^4=(c^8+x^8)+c^2s^2(4c^4+6c^2s^2+4s^4)=$$ $$=frac 38+c^2s^2(4(c^2+s^2)^2-2c^2s^2)=$$ $$=frac 38+c^2s^2(4-2c^2s^2)iff$$ $$iff(0leq pleq 1land frac 516=2p-p^2)$$ $$iff p=1- sqrt 11;/4iff$$ $$iff |sin 2x|=sqrt 4p=sqrt 4-sqrt 11.$$

Note: The 2nd, 3rd, & 4th displayed lines are a single sentence that is true iff $c^8+x^8=frac 38.$


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edited Sep 9, 2018 at 16:16
answered Sep 9, 2018 at 16:10
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DanielWainfleetDanielWainfleet
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eginalignsin^8 x+cos^8 x&=(sin^4 x+cos^4)^2-2sin^4 xcos^4 x\&=left<(sin^2 x+cos^2 x)^2-2sin^2 xcos^2 x ight>^2-cfrac sin^4 2x 8\&=(1-cfrac sin^2 2x2)^2--cfrac sin^4 2x 8endalignLet $t=sin^2 2x$, we have$$(1-cfrac t2)^2-cfrac t^28=cfrac 48128Leftrightarrow t^2-8t+5=0,$$ which implies$$sin^2 2x=4-sqrt11 hspace1in ext(with the other invalid root discarded)$$

hence $$ x=pm cfrac sin^-1(sqrt4-sqrt11)2$$or more precisely,$$x=pmcfrac sin^-1(sqrt4-sqrt11)2+kpi,>kin ingamemobi.combbZ$$


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edited Nov 6, 2018 at 17:51
answered Nov 6, 2018 at 15:38
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LanceLance
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Let $c=cosx$ & $s=sinx$ :

$s^8+c^8= frac482^7*frac22$

$s^8+c^8= frac962^8 $

$(fracG2^4)^2+(fracH2^4)^2=frac962^8implies G^2+H^2=96 $

$implies s=pmfracsqrt<4>G2 wedge c=pmfracsqrt<4>96-G^22$

We gonna use: $s^2+c^2=1 ightarrow c^2=1-s^2$:

$fracsqrt96-G^24=1-fracsqrtG4$

$sqrt96-G^2=4-sqrtG ightarrow$ WolframAlpha finds the solution : $Gapprox9.7587$

For $sqrt96-g^4=4-g$ we get real solutions : $g_1approx0.0406-e vee g_2approxpi-0.0177;$ cause $g_1^2 eq Gimplies g_2$ gonna be our result of $g^2=G$.$(gapprox 3.12389354026129)$

So:

$s=pmfracsqrtg2$

We consider for $sinx$ in the 1st quarter:

$fracsqrt32Direct:

$x=arcsin(fracsqrtg2)approx62.1^circ$

History và altenative version (with different* solution):

All starts with:

$(c^4+is^4)(c^4-is^4)=frac62^4$

$(fracA2^2)^2+(fracB2^2)^2=frac62^4implies A^2+B^2=6 $

$implies c^4=fracA4 wedge s^4=fracsqrt6-A^24$

Altenative for:$c=pmsqrt<4>fracA4 wedge s=pmsqrt<4>fracB4$ with $s^2+c^2=1$ :

$sqrtfracB4=1-sqrtfracA4$

$6-A^2=<2-sqrtA>^2$

$6-A^2=4-2sqrtA+A$

$0=A^2+A-2sqrtA-2 ightarrow$ WolframAlpha finds the solution : $Aapprox1.7049$

So:

$c=pmfracsqrt<4>Asqrt2$

We consider for $cosx$ in the 1st quarter:

$x=arccos(fracsqrt<4>Asqrt2)approx36.1^circ$

*Also: If I would stand by $s=pmfracsqrt<4>Asqrt2 ightarrow xapprox53.9^circ$ (1st quarter)