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Question Number 20036 by mondodotto@gmail.com last updated on 20/Aug/17

Answered by sma3l2996 last updated on 21/Aug/17

2sin^(−1) (x(√6))=(π/2)−sin^(−1) (4x)⇔sin(2sin^(−1) (x(√6)))=sin((π/2)−sin^(−1) (4x)) 2sin(sin^(−1) (x(√6)))cos(sin^(−1) (x(√6)))=cos(sin^(−1) (4x)) 2x(√6)(√(1−6x^2 ))=(√(1−16x^2 )) 24x^2 (1−6x^2 )=1−16x^2 24x^2 −144x^4 +16x^2 =1 144x^4 −40x^2 =−1⇔144x^4 −2×(5/3)×12x^2 +((25)/9)=((25)/9)−1 (12x^2 −(5/3))^2 =((16)/9)⇔x^2 =(1/4) or x^2 =(1/(36)) x=(1/2);((−1)/2);(1/6);((−1)/6)

$$\mathrm{2}{sin}^{−\mathrm{1}} \left({x}\sqrt{\mathrm{6}}\right)=\frac{\pi}{\mathrm{2}}−{sin}^{−\mathrm{1}} \left(\mathrm{4}{x}\right)\Leftrightarrow{sin}\left(\mathrm{2}{sin}^{−\mathrm{1}} \left({x}\sqrt{\mathrm{6}}\right)\right)={sin}\left(\frac{\pi}{\mathrm{2}}−{sin}^{−\mathrm{1}} \left(\mathrm{4}{x}\right)\right) \\ $$$$\mathrm{2}{sin}\left({sin}^{−\mathrm{1}} \left({x}\sqrt{\mathrm{6}}\right)\right){cos}\left({sin}^{−\mathrm{1}} \left({x}\sqrt{\mathrm{6}}\right)\right)={cos}\left({sin}^{−\mathrm{1}} \left(\mathrm{4}{x}\right)\right) \\ $$$$\mathrm{2}{x}\sqrt{\mathrm{6}}\sqrt{\mathrm{1}−\mathrm{6}{x}^{\mathrm{2}} }=\sqrt{\mathrm{1}−\mathrm{16}{x}^{\mathrm{2}} } \\ $$$$\mathrm{24}{x}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{6}{x}^{\mathrm{2}} \right)=\mathrm{1}−\mathrm{16}{x}^{\mathrm{2}} \\ $$$$\mathrm{24}{x}^{\mathrm{2}} −\mathrm{144}{x}^{\mathrm{4}} +\mathrm{16}{x}^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{144}{x}^{\mathrm{4}} −\mathrm{40}{x}^{\mathrm{2}} =−\mathrm{1}\Leftrightarrow\mathrm{144}{x}^{\mathrm{4}} −\mathrm{2}×\frac{\mathrm{5}}{\mathrm{3}}×\mathrm{12}{x}^{\mathrm{2}} +\frac{\mathrm{25}}{\mathrm{9}}=\frac{\mathrm{25}}{\mathrm{9}}−\mathrm{1} \\ $$$$\left(\mathrm{12}{x}^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{3}}\right)^{\mathrm{2}} =\frac{\mathrm{16}}{\mathrm{9}}\Leftrightarrow{x}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\:\:{or}\:\:{x}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{36}} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}};\frac{−\mathrm{1}}{\mathrm{2}};\frac{\mathrm{1}}{\mathrm{6}};\frac{−\mathrm{1}}{\mathrm{6}} \\ $$

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