Respuesta :

lukyo
[tex]\cos^2 x\sin x-3\sin x=0\\\\ \sin x\cdot (\cos^2 x-3)=0\\\\ \begin{array}{rcl} \sin x=0&~\text{ or }~&\cos^2 x - 3=0\\\\ \sin x=0&~\text{ or }~&\cos^2 x=3\\\\ \sin x=0&~\text{ or }~&\cos x=\pm \sqrt{3} \end{array}[/tex]


Since [tex]|\cos x|\le 1,[/tex] the 2nd equation has no solutions for x. So

[tex]\sin x=0\\\\ x=k\pi~~~~\text{where }k\text{ is an integer.}[/tex]


Solution: [tex]S=\{x:~x=k\pi,~k\in\mathbb{Z}\}[/tex]