The equation ${\sin }^{4}x+a{\sin }^{2}x+1=0$  will have a solution if $a$  belongs to

The equation ${\sin }^{4}x+a{\sin }^{2}x+1=0$

Let $u={\sin }^{2}x$ , then $u\in \left[0,1\right]$

$\therefore$  $u^2+au+1=0$  should have at least one solution in $\left[0,1\right]$

$\therefore$  discriminant$=a^2-4\ge 0$  $\Rightarrow a\le -2$   or   $a\ge 2$

Product of roots$=1,$  only one root may lie in $\left[0,1\right]$

$\therefore$  $f\left(0\right)f\left(1\right)\le 0$ $\Rightarrow a+2\le 0\Rightarrow a\le -2$

Hence $a\in \left(-\infty ,-2\right]$