Question

If the maximum value of $x$ which satisfies the inequality ${\left(sin\right)}^{-1}\left(sinx\right)\ge {\left(cos\right)}^{-1}\left(sinx\right)$ 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

Answer: 1.57

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