Question

If $x\in [0,100\pi ]$, then the number of solutions of the equation ${\sin }^{4}x-{\cos }^{2}x\sin x+2{\sin }^{2}x+\sin x=0$ is

Answer 1

${\sin }^{4}x-{\cos }^{2}x\sin x+2{\sin }^{2}x+\sin x=0$
$\Rightarrow \sin x({\sin }^{3}x-{\cos }^{2}x+2\sin x+1)=0$
$\Rightarrow \sin x({\sin }^{3}x+{\sin }^{2}x+2\sin x)=0$
$\Rightarrow {\sin }^{2}x({\sin }^{2}x+\sin x+2)=0$
$\Rightarrow \sin x=0\text{or}({\sin }^{2}x+\sin x+2)=0$
When ${\sin }^{2}x+\sin x+2=0$
$D=1-8=-7<0$
So, no real solution.
Therefore,
$\sin x=0\Rightarrow x=n\pi$
We know that $x\in [0,100\pi ]$,
$\therefore x=0,\pi ,2\pi ,...,100\pi$

Hence, the number of solutions is $101$.