An Interesting Trigonometric Equation
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Күн бұрын
@xualain3129 Ай бұрын
If both sin(x) and cos(x) are not equal to zero, then sin(x)^5 < sin(x)^2 …..(1) cos(x)^7
@StaR-uw3dc Ай бұрын
nice solution but from the equation sin²x(sin³x-1)=cos²x(1-cos⁵x) we can get the system sin³x-1=0 AND cos²x=0 OR 1-cos⁵x=0 AND sin²x=0. This system has the solution as in video.
@SidneiMV Ай бұрын
awesome!
@ShortsOfSyber Ай бұрын
@@StaR-uw3dc thanks. This is pretty good
@georgesbv1 Ай бұрын
I wanted to say the same thing.
@RyanLewis-Johnson-wq6xs Ай бұрын
Sin(x)=1 X=π/2,(5π)/2, Cos(x)=1 x=0,2π
@giuseppemalaguti435 Ай бұрын
Dalle fattorizzazioni si trova sia (1-sinx)=0...(1-cosx)=0...
@RealQinnMalloryu4 Ай бұрын
{sin^5+sin^ ➖}+{ x+x ➖ }+{cos^7+cos^7 ➖ }+{x+x ➖ }={sin^10+x^2}+{cos^14+x^2}={sin^x^12+cosx^16}=sincosx^28 sincosx^12^16 sincosx^2^10^2^14 sincosx^1^2^5^1^7^7.sincosx^2^2^3^3^4^3^4 sincosx^1^1^1^1^2^23^2^2 sincosx^1^1^3^1^2 sincosx^3^2 (sincos x ➖ 3sincosx+2).
@RyanLewis-Johnson-wq6xs Ай бұрын
(Sin[x])^5+(Cos[x])^7=1 X=0 X=2πn, n element Z
@jamescollis7650 Ай бұрын
...but are there any other complex solutions? :)
@ВиталийТруш-н1р Ай бұрын
t=tan(x/2) sinx=2t/(1+t^2) cosx=(1-t^2)/(1+t^2) (2t/(1+t^2))^5+((1-t^2)/(1+t^2))^7=1 (2t)^5*(1+t^2)^2+(1-t^2)^7=(1+t^2)^7 and 1+t^2 not =0 32t^5*(1+t^2)^2+1-7t^2+21t^4-35t^6+35t^8-21t^10+7t^12-t^14=1+7t^2+21t^4+35t^6+35t^8+21t^10+7t^12+t^14 32t^5*(1+t^2)^2-14t^2-70t^6-42t^10-2t^14=0 t=0 or 16t^3*(1+2t^2+t^4)-7-35*t^4-21t^8-t^12=0 This equation has root t=1. We can divide to (t-1)
@ВиталийТруш-н1р Ай бұрын
Result (calculate online) -x^11-x^10-x^9-x^8-22∙x^7-6∙x^6-6∙x^5+26∙x^4-9∙x^3+7∙x^2+7∙x+7=0 Instead x t. t=1 is root of this equation Divide to (t-1)
@ВиталийТруш-н1р Ай бұрын
-x^10-2∙x^9-3∙x^8-4∙x^7-26∙x^6-32∙x^5-38∙x^4-12∙x^3-21∙x^2-14∙x-7=0. Instead x t. t=1 is not root of this equation. Initial equation has not other real roots except sinx=0 and sinx=1 (2t/(1+t^2) =0 or 1, t= 0 or 1). This equation has complex roots, but find it...
@jamescollis7650 Ай бұрын
@@ВиталийТруш-н1р Very clever :) So there are more complex solutions!
@michaelyap939 Ай бұрын
Why go so long ? Why not just sin²x(sin³x-1)=cos²x(1-cos⁵x) ?