Question

Given $\frac{1}{2}\le x\le 1$, then find:
$co{s}^{-1}x+co{s}^{-1}\frac{x}{2}+\sqrt{\frac{(3-3x^2)}{2}}$

Answer 1

$co{s}^{-1}x+co{s}^{-1}(\frac{x}{2}+\frac{\sqrt{3-3x^2}}{2})=\frac{\pi }{3}$
Equate:
$x=\frac{x}{2}+\frac{\sqrt{3-3x^2}}{2}$
$2x=x+\sqrt{3-3x^2}$
$x=\sqrt{3-3x^2}$
squaring both sides:
$x^2=3-3x^2$
$4x^2=3$
$x^2=\frac{3}{4}$
$x=\pm \frac{\sqrt{3}}{2}$
Now putting in original equation, we do not take $-\frac{\sqrt{3}}{2}$ since it will not give the desired value:
$co{s}^{-1}\left(\frac{\sqrt{3}}{2}\right)+co{s}^{-1}(\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{2}\left(\sqrt{1-\frac{3}{4}}\right))$
$co{s}^{-1}\left(\frac{\sqrt{3}}{2}\right)+co{s}^{-1}(\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4})$
$2co{s}^{-1}\left(\frac{\sqrt{3}}{2}\right)=2\times \frac{\pi }{6}=\frac{\pi }{3}$
hence $x=\frac{\sqrt{3}}{2}$