Question
If the solutions for $$\theta $$ from the equation $${\sin ^2}\theta - 2\sin \theta + \lambda = 0$$ lie in \[\begin{array}{*{20}{c}}
\cup \\
{n \in {\Bbb Z}}
\end{array}\left( {2n\pi - \frac{\pi }{6},\overline {2n + 1} \,\pi + \frac{\pi }{6}} \right)\] then the set of possible values of $$\lambda $$ is
A.
$$\left( { - \frac{5}{4},1} \right]$$
B.
$$\left( { - \infty ,1} \right]$$
C.
$$\left( { - \frac{5}{4}, + \infty } \right]$$
D.
$$\left\{ 1 \right\}$$
Answer :
$$\left( { - \frac{5}{4},1} \right]$$
Solution :
$$\sin \theta = \frac{{2 \pm \sqrt {4 - 4\lambda } }}{2} = 1 \pm \sqrt {1 - \lambda } .$$ For real values, $$1 - \lambda \geqslant 0,\,\,{\text{i}}{\text{.e}}{\text{., }}\lambda \leqslant 1.$$
As $$ - 1 \leqslant \sin\theta \leqslant 1,\sin\theta = 1 - \sqrt {1 - \lambda } .$$
From the question, $$\sin\theta > - \frac{1}{2}.\,{\text{Thus }} - \frac{1}{2} < 1 - \sqrt {1 - \lambda } \leqslant 1$$
$$\eqalign{
& {\text{or, }} - \frac{3}{2} < - \sqrt {1 - \lambda } \leqslant 0 \cr
& \Rightarrow \,\,\sqrt {1 - \lambda } < \frac{3}{2} \cr
& \Rightarrow \,\,1 - \lambda < \frac{9}{4} \cr
& \Rightarrow \,\,\lambda > - \frac{5}{4}. \cr} $$