cos^(−1) ((2x^2 −1)/(2x^2 )) + cos^(−1) ((x^2 −2)/x^2 )=120°=((2π)/3) Find x


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Question Number 61511 by Kunal12588 last updated on 03/Jun/19

${c o s}^{- 1} \frac{2 x^{2} - 1}{2 x^{2}} + {c o s}^{- 1} \frac{x^{2} - 2}{x^{2}} = 120 ° = \frac{2 π}{3}$ $Find x$
	Answered by MJS last updated on 03/Jun/19

	$\cos^{- 1} \frac{2 x^{2} - 1}{2 x^{2}} + \cos^{- 1} \frac{x^{2} - 2}{x^{2}} = \frac{2 π}{3}$ $π - \sin^{- 1} \frac{2 x^{2} - 1}{x^{2}} - \sin^{- 1} \frac{x^{2} - 2}{x^{2}} = \frac{2 π}{3}$ $\sin^{- 1} \frac{2 x^{2} - 1}{x^{2}} = \frac{π}{3} - \sin^{- 1} \frac{x^{2} - 2}{x^{2}}$ $\frac{2 x^{2} - 1}{2 x^{2}} = - \frac{x^{2} - 2 - \sqrt{3 (x^{2} - 1)}}{2 x^{2}} [beware of false solutions!!!]$ $\Rightarrow x = \pm 1 [false!] \lor x = \pm \frac{\sqrt{21}}{3}$
	Commented by Kunal12588 last updated on 04/Jun/19

	$t h a n k y o u s i r$
	Commented by Kunal12588 last updated on 04/Jun/19

	$h o w d i d y o u k n o w x = \pm 1 i s f a l s e$
	Commented by Kunal12588 last updated on 04/Jun/19

	$\frac{2 x^{2} - 1}{2 x^{2}} = - \frac{x^{2} - 2 - 2 \sqrt{3 (x^{2} - 1)}}{2 x^{2}} s i r$
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