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Q.
The equation $\sin^{4} x - \left(k+2\right)\sin^{2} x -\left(k + 3\right) =0 $ possesses a solution if
Trigonometric Functions
Solution:
We have,$\sin^{4} x - \left(k+2\right)\sin^{2} x -\left(k + 3\right) =0$
$ \Rightarrow \sin^{2} x = \frac{\left(k+2\right) \pm \sqrt{\left(k+2\right)^{2} + 4 \left(k+3\right)}}{2} $
$ = \frac{\left(k+2\right)\pm\left(k+4\right)}{2}$
$ \Rightarrow \sin^{2} x = k + 3$
$ (\because \sin^{2} x = - 1$ is not possible)
Since $ 0 \le\sin^{2} x \le1 , $
$ \therefore 0 \le k + 3\le1 or -3 \le k \le - 2 $