Question

Solve for x $co{s}^{-1}x+si{n}^{-1}\frac{x}{2}=\frac{\pi }{6}$

Answer 1

$co{s}^{-1}x+si{n}^{-1}\frac{x}{2}=\frac{\pi }{6}\frac{\pi }{2}-si{n}^{-1}x+si{n}^{-1}\frac{x}{2}=\frac{\pi }{6}si{n}^{-1}=si{n}^{-1}\frac{x}{2}=\frac{\pi }{2}=\frac{\pi }{6}=\frac{2\pi }{6}si{n}^{-1}x=si{n}^{-1}\frac{x}{2}+si{n}^{-1}\left(\frac{\sqrt{3}}{1}\right)=si{n}^{-1}\left(\frac{x}{2}\sqrt{1-(\frac{\sqrt{3}}{2}{)}^2}\right)+\frac{\sqrt{3}}{2}\sqrt{1-(\frac{x}{2}{)}^2}=si{n}^{-1}\left(\frac{x}{2}\right(\frac{1}{2})+\frac{\sqrt{3}}{2}\frac{\sqrt{4-x^2}}{2})=si{n}^{-1}x=si{n}^{-1}(\frac{x}{4}+\frac{\sqrt{3}\sqrt{4-x^2}}{4})\Rightarrow x=\frac{x}{4}+\frac{\sqrt{3(4-x{)}^2}}{4}x=\frac{x}{4}=\frac{\sqrt{3(4-x^2)}}{4}\frac{3x}{4}=\frac{\sqrt{3(4-x^2)}}{4}9x^2=3(4-x^2)9x^2=12-3x^{2}12x^2=12x^2=1x=\pm 1$