cos-1-2x-cos-1-x-pi-6-find-x-

Question Number 105650 by bemath last updated on 30/Jul/20

\cos^{- 1} (2 x) + \cos^{- 1} (x) = \frac{π}{6}

f i n d x

Answered by bramlex last updated on 30/Jul/20

\cos^{- 1} (2 x) = \frac{π}{6} - \cos^{- 1} (x)

\cos (\cos^{- 1} (2 x)) = \cos (\frac{π}{6} - \cos^{- 1} (x))

2 x = \frac{\sqrt{3}}{2} x + \frac{1}{2} \sin (\cos^{- 1} (x))

4 x = \sqrt{3} x + \sin (\cos^{- 1} (x))

x (4 - \sqrt{3}) = \sin (\cos^{- 1} (x))

x (4 - \sqrt{3}) = \sqrt{1 - \cos^{2} (\cos^{- 1} (x))}

x (4 - \sqrt{3}) = \sqrt{1 - x^{2}}

x^{2} {(4 - \sqrt{3})}^{2} = 1 - x^{2}

x^{2} (19 - 8 \sqrt{3} + 1) = 1

x^{2} = \frac{1}{20 - 8 \sqrt{3}} = \frac{1}{4 (5 - 2 \sqrt{3})}

x = \pm \frac{1}{2 \sqrt{5 - 2 \sqrt{3}}} . ▴

Answered by mathmax by abdo last updated on 31/Jul/20

\Rightarrow \cos (arcos (2 x) + cosx) = \frac{\sqrt{3}}{2} \Rightarrow

2 x . x - \sin (arcos (2 x)) \sin (cosx) = \frac{\sqrt{3}}{2} \Rightarrow

{2 x}^{2} - \sqrt{1 - {(2 x)}^{2}} \times \sqrt{1 - x^{2}} = \frac{\sqrt{3}}{2} \Rightarrow

({2 x}^{2} - \frac{\sqrt{3}}{2}) = \sqrt{1 - {4 x}^{2}} \times \sqrt{1 - x^{2}} \Rightarrow

{({2 x}^{2} - \frac{\sqrt{3}}{2})}^{2} = (1 - {4 x}^{2}) (1 - x^{2}) \Rightarrow

{4 x}^{4} - 2 \sqrt{3} x^{2} + \frac{3}{4} = 1 - x^{2} - {4 x}^{2} + {4 x}^{4} \Rightarrow

- 2 \sqrt{3} x^{2} + \frac{3}{4} = 1 - {5 x}^{2} \Rightarrow - 8 \sqrt{3} x^{2} + 3 = 4 - {20 x}^{2} \Rightarrow

(20 - 8 \sqrt{3}) x^{2} = - 3 \Rightarrow (8 \sqrt{3} - 20) x^{2} = 3 \Rightarrow x^{2} = \frac{3}{8 \sqrt{3} - 20} \Rightarrow

x = \bar{+} \sqrt{\frac{3}{8 \sqrt{3} - 20}}