Question

If the maximum value of $x$ which satisfies the inequality $\left(sin\right)^{- 1} \left(sin ⁡ x\right)\geq \left(cos\right)^{- 1} ⁡ \left(sin ⁡ x\right)$ for $x\in \left(\frac{\pi }{2} , 2 \pi \right)$ is $\lambda $ , then $\frac{2 \lambda }{3}$ is equal to (take $\pi =3.14$ )

Answer 1

Let $t=sin x$
From the graphs of $sin^{- 1} t$ and $cos^{- 1} t$ , we know that,
$sin^{- 1} t\geq cos^{- 1} ⁡ t$ for $t\in \left[\frac{1}{\sqrt{2}} , 1\right]$
$\Rightarrow sin x\in \left[\frac{1}{\sqrt{2}} , 1\right]$
Hence, $x \in\left(\frac{\pi}{2}, \frac{3 \pi}{4}\right]$
Maximum value is $\frac{3 \pi }{4}=\lambda $
So, $\frac{2 \lambda }{3}=\frac{\pi }{2}=1.57$