Question

The solution of the equation $2{\cos }^{-1}x={\sin }^{-1}\left(2x\sqrt{1-x^2}\right)$

• A. $[-1,0]$
• B. $[0,1]$
• C. $[-1,1]$
• D. $[\frac{1}{\sqrt{2}},1]$

Answer 1

The correct option is D $[\frac{1}{\sqrt{2}},1]$

$2x\sqrt{1-x^2}\le |1|$
$4x^2(1-x^2)\le 1$
$4x^2-4x^4\le 1$
$4x^4-4x^2+1\ge 0$
$(2x^2-1{)}^2\ge 0$
$2x^2-1\ge 0$
$x^2\ge \frac{1}{2}$
$xϵ[-1,-\frac{1}{\sqrt{2}}]\cup [\frac{1}{\sqrt{2}},1]$ ...(ii)
Hence the range of solution for the above expression for the given value of x will be
However,
Range for $co{s}^{-1}\left(x\right)=[0,\pi ]$
$si{n}^{-1}\left(x\right)=[\frac{-\pi }{2},\frac{\pi }{2}]$
Hence the common part is $[0,\frac{\pi }{2}]$, ...(i)
Putting $x=cos\theta$
$\theta ϵ[0,\frac{\pi }{2}]$ ...(iii) ... from ii.
Hence from i, ii, and iii, we get
$x=[\frac{1}{\sqrt{2}},1]$