Question

Resolver senx.cos^2x=senx
(Es coseno al cuadrado) ayuda porta

Answer 1

Tenemos:

$\sin \left(x\right)\cos ^2\left(x\right)=\sin \left(x\right)\\\\\sin \left(x\right)\cos ^2\left(x\right)-\sin \left(x\right)=0\\\\\sin \left(x\right)\left(\cos \left(x\right)+1\right)\left(\cos \left(x\right)-1\right)=0\\\\\sin \left(x\right)=0\quad \mathrm{o}\quad \cos \left(x\right)+1=0\quad \mathrm{o}\quad \cos \left(x\right)-1=0$

sin(x) = 0:

$\sin \left(x\right)=0\quad\\\\x=0+2\pi n,\:x=\pi +2\pi n\\\\x=0+2\pi n\\\\x=2\pi n$

Respuesta final:

$x=2\pi n,\:x=\pi +2\pi n$

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