Question

If the solutions for $\theta$ from the equation ${\sin }^2\theta -2\sin \theta +\lambda =0$ lie in $\bigcup _{nϵz}(2n\pi -\frac{\pi }{6},(2n+1)\pi +\frac{\pi }{6})$ then the set of possible values of $\lambda$ is

• A. $(-\frac{5}{4},1]$
• B. $(-\infty ,1]$
• C. $(-\frac{5}{4},+\infty ]$
• D. $\left\{1\right\}$

Answer 1

The correct option is A $(-\frac{5}{4},1]$

$si{n}^2\theta -2sin\theta +\lambda =0$

$\Rightarrow \lambda =-(si{n}^2\theta -2sin\theta +1-1)$

$\Rightarrow \lambda =1-(sin\theta -1{)}^2$

Given $\theta \in \bigcup (2n\pi -\frac{\pi }{6},(2n+1)\pi -\frac{\pi }{6})$

$\Rightarrow -\frac{1}{2}\le sin\theta \le 1$

thus range of $\lambda$ so that equation posses solution is,

$\lambda ϵ(-\frac{5}{4},1]$

Hence, option 'A' is correct.