Rozwiązane

sin³x - 4sin²x - sinx = -4

Jak to rozwiązać?

Odpowiedź :

Konrad509viz

[tex]\sin^3x-4\sin^2x-\sin x=-4\\\sin^3x-4\sin^2x-\sin x+4=0\\\sin^2 x(\sin x-4)-1(\sin x-4)=0\\(\sin^2 x-1)(\sin x-4)=0\\(\sin x-1)(\sin x+1)(\sin x-4)=0\\\\\sin x-1=0\\\sin x=1\\x=\dfrac{\pi}{2}+2k\pi\\\\\sin x+1=0\\\sin x=-1\\x=-\dfrac{\pi}{2}+2k\pi\\\\\sin x-4=0\\\sin x=4\\\x\in\emptyset\\\\\boxed{x\in\left\{-\dfrac{\pi}{2}+2k\pi,\dfrac{\pi}{2}+2k\pi\right\}}\\\\(k\in\mathbb{Z})[/tex]

Louie314viz

Rozwiązanie:

Równanie:

[tex]sin^{3}x-4sin^{2}x-sinx+4=0[/tex]

Podstawmy [tex]t=sinx[/tex], gdzie [tex]-1\leq t\leq 1[/tex] :

[tex]t^{3}-4t^{2}-t+4=0[/tex]

[tex]t^{2}(t-4)-(t-4)=0[/tex]

[tex](t-4)(t^{2}-1)=0[/tex]

[tex](t-4)(t-1)(t+1)=0[/tex]

[tex]t=4 \notin D \vee t=-1 \in D \vee t=1 \in D[/tex]

Zatem:

[tex]sinx=1 \vee sinx=-1[/tex]

[tex]$x=-\frac{\pi}{2} +2k\pi \vee x=\frac{\pi }{2}+2k\pi[/tex]

Viz Inne Pytanie