The equation $\sin^4x-(k+2)\sin^2x-(k+3)=0$ possesses a solution if: The equation $\sin^4x-(k+2)\sin^2x-(k+3)=0$ possesses a solution if :
$A)\: k>-3 $
$B)\: k<-2 $
$C)\: -3\le k\le-2  $
$D)$:  $k$ is any (+ve) value
My attempt is as follows:-
$$\sin^2(x_1)\cdot\sin^2(x_2)=-(k+3)$$
Here $\sin^2(x_1),\sin^2(x_2)$ signify the solution of given quadratic in $\sin^2x$
$$-(k+3)\ge0$$
$$k+3\le0$$
$$k\le-3$$
$$\dfrac{\sin^2(x_1)+\sin^2(x_2)}{2}\ge0$$
$$(k+2)\ge0$$
$$k\ge-2$$
So seems like there is no solution but actual answer is $C)$. What am I missing here?
 A: $$\sin ^4 x-(k+2)\sin^2 x-(k+3)=0~~~(1)$$
Let $sin^2x =z$
$$f(z)=z^2-(k+2)z-(k+3)=0~~~(2)$$ there are two cases
note thta for this quadratic (2) has $B^2 -4AC =(k+4)^2 \ge 0$
Case-1: exactly one root should be in $(0,1) \implies f(0)f(1)<0 \implies -3<k<-2$.
Case-2: Both roots in  $(0,1) \implies 0<z_0=(k+2)/2<1 ~and ~ f(1)>1 ~and~ f(0)>0.$
In this case we get $ -2 <k<0, ~and~ k<-3 ~and~ k<-2$ whose overlap is null.
So only case (1) is possible. Finally Eq. (1) will exactly one solution if $$-3<k<-2$$ which is option (C).
e
A: There is a flaw in this step :
$$\sin^2(x_1)\cdot\sin^2(x_2)=-(k+3)$$
You are assuming that both the solution of quadratic equation in $\sin^2(x)$ are feasible, which is not necessary for solution to exist as it suffices to have one solution feasible.
For example : If you choose $k=-2$, you get $\sin^2(x)=\pm 1$, in which only one solution is feasible.
Hint : For solving, directly compute the root of the quadratic in terms of $k$ and find those value of $k$ for which at least one of them lies between 0 and 1.
A: Hint:
The quadratic equation in $\sin^2x$ always has real roots since its discriminant is 
$$\Delta=(k+2)^2+4(k+3)=(k+4)^2,$$
which is nonnegative.
Also, it must have at least one (real) root between $0$ and $1$. Examine the different possible configurations. Your life will be a lot simpler if you remember that a real number $\alpha$ separates the (real) roots of a quadratic polynomial $p(x)=ax^2+bx+c\:$ if and only if $ap(\alpha)<0$.
