Question

If the equation ${\sin }^{4}x-(k+2){\sin }^{2}x-(k+3)=0$ has a solution then $k$ must lie in the interval:

Answer 1

$si{n}^{4}x-(k+2)si{n}^{2}x-(k+3)=0$

$si{n}^{2}x\in [0,1]$

put $si{n}^{2}x=t,t\in [0,1]$

$t^2-(k+2)t-(k+3)=0$

put $t=0$

$0-(k+2)0-k-3=0$

$k=-3$

put $t=1$

$1-k-2-k-3=0$

$\Rightarrow -2k=4$

$k=-2$

$k\in [-3,-2]$