Question

If $co{s}^{-1}\left(\sqrt{16}x\right)+co{s}^{-1}\left(3\sqrt{3}{x}^2\right)=\frac{\pi }{2}$, then x has:

• A. Only one value
• B. Two values
• C. three values
• D. four values

Answer 1

The correct option is A Only one value

Clearly $0\le x\le 1$ as for x < 0, $co{s}^{-1}\left(\sqrt{6}x\right)\in Q_2$
$co{s}^{-1}\left(\sqrt{16}x\right)+co{s}^{-1}\left(3\sqrt{3}{x}^2\right)=\frac{\pi }{2}\Rightarrow co{s}^{-1}\left(3\sqrt{3}{x}^2\right)=\frac{\pi }{2}-co{s}^{-1}\left(\sqrt{16}x\right)=si{n}^{-1}\left(\sqrt{16}x\right)\Rightarrow 3\sqrt{3}{x}^2=\sqrt{1-6x^2}\Rightarrow 27x^4+6x^2-1=0\Rightarrow (9x^2-1)(3x^2+1)=0\Rightarrow x^2=\frac{1}{9}\Rightarrow x=\frac{1}{3}(\because x>0)$