Question

Find real value of x,if
${\cos }^{-1}\left(\sqrt{6}x\right)+{\cos }^{-1}\left(3\sqrt{3}{x}^2\right)=\frac{\pi }{2}$.

• A. $x=\pm \frac{1}{3}$
• B. $x=\pm \frac{1}{2}$
• C. $x=\pm \frac{1}{\sqrt{3}}$
• D. $x=\pm \frac{1}{\sqrt{2}}$

Answer 1

The correct option is A $x=\pm \frac{1}{3}$

${\cos }^{-1}\left(\sqrt{6x}\right)+{\cos }^{-1}\left(3\sqrt{3}{x}^2\right)=\frac{\pi }{2}$ ....(2)

${\cos }^{-1}\left(3\sqrt{3}{x}^2\right)={\sin }^{-1}\left(\sqrt{6}x\right)$

${\cos }^{-1}\left(3\sqrt{3}{x}^2\right)={\cos }^{-1}\left(\sqrt{1-6x^2}\right)$

constration on x's $1-6x^2>0$

$\frac{-16}{\sqrt{6}}x\le \frac{1}{16}$

taking call both sides
$3\sqrt{3}{x}^2=\sqrt{1-6x^2}$

$27x^4+6x^2-1=0$ ...(1)

solving (1) $\Rightarrow x=\pm \frac{1}{3}$ ... (3)

Cross checking roots in (3) with (2) we get the only roots $\frac{1}{8}$ as in $co{s}^{-1}\left(\sqrt{6}x\right)>\frac{\pi }{2}$ and this is not possible as $\pi >{\cos }^{-1}\left(x\right)>0$.