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

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

$${cos}^{−\mathrm{1}} \:\frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }\:+\:{cos}^{−\mathrm{1}} \:\frac{{x}^{\mathrm{2}} −\mathrm{2}}{{x}^{\mathrm{2}} }=\mathrm{120}°=\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$$\mathrm{Find}\:\:{x} \\ $$

Answered by MJS last updated on 03/Jun/19

cos^(−1) ((2x^2 −1)/(2x^2 )) +cos^(−1) ((x^2 −2)/x^2 ) =((2π)/3) π−sin^(−1) ((2x^2 −1)/x^2 ) −sin^(−1) ((x^2 −2)/x^2 )=((2π)/3) sin^(−1) ((2x^2 −1)/x^2 ) =(π/3)−sin^(−1) ((x^2 −2)/x^2 ) ((2x^2 −1)/(2x^2 ))=−((x^2 −2−(√(3(x^2 −1))))/(2x^2 )) [beware of false solutions!!!] ⇒ x=±1 [false!] ∨ x=±((√(21))/3)

$$\mathrm{cos}^{−\mathrm{1}} \:\frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }\:+\mathrm{cos}^{−\mathrm{1}} \:\frac{{x}^{\mathrm{2}} −\mathrm{2}}{{x}^{\mathrm{2}} }\:=\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$$\pi−\mathrm{sin}^{−\mathrm{1}} \:\frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} }\:−\mathrm{sin}^{−\mathrm{1}} \:\frac{{x}^{\mathrm{2}} −\mathrm{2}}{{x}^{\mathrm{2}} }=\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$$\mathrm{sin}^{−\mathrm{1}} \:\frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} }\:=\frac{\pi}{\mathrm{3}}−\mathrm{sin}^{−\mathrm{1}} \:\frac{{x}^{\mathrm{2}} −\mathrm{2}}{{x}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }=−\frac{{x}^{\mathrm{2}} −\mathrm{2}−\sqrt{\mathrm{3}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}}{\mathrm{2}{x}^{\mathrm{2}} }\:\left[\mathrm{beware}\:\mathrm{of}\:\mathrm{false}\:\mathrm{solutions}!!!\right] \\ $$$$\Rightarrow\:{x}=\pm\mathrm{1}\:\left[\mathrm{false}!\right]\:\vee\:{x}=\pm\frac{\sqrt{\mathrm{21}}}{\mathrm{3}} \\ $$

Commented by Kunal12588 last updated on 04/Jun/19

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Commented by Kunal12588 last updated on 04/Jun/19

how did you know x=±1 is false

$${how}\:{did}\:{you}\:{know}\:{x}=\pm\mathrm{1}\:{is}\:{false} \\ $$

Commented by Kunal12588 last updated on 04/Jun/19

((2x^2 −1)/(2x^2 ))=−((x^2 −2−2(√(3(x^2 −1))))/(2x^2 )) sir

$$\frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }=−\frac{{x}^{\mathrm{2}} −\mathrm{2}−\mathrm{2}\sqrt{\mathrm{3}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}}{\mathrm{2}{x}^{\mathrm{2}} }\:{sir} \\ $$

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