Question-21297

Question Number 21297 by NECx last updated on 19/Sep/17

Answered by sma3l2996 last updated on 20/Sep/17

2sin^(−1) (x(√6))+sin^(−1) (4x)=(π/2) sin^(−1) (4x)=(π/2)−2sin^(−1) (x(√6)) sin(sin^(−1) (4x))=sin((π/2)−2sin^(−1) (x(√6))) we have sin(π/2−a)=cosa so : 4x=cos(2sin^(−1) (x(√6))=1−2sin^2 (sin^(−1) (x(√6))) 4x=1−2×(x(√6))^2 ⇔12x^2 +4x−1=0 x^2 +(1/3)x−(1/(12))=0⇔x^2 +2.(1/6)x+(1/(36))=(1/(36))+(1/(12))=(1/9) (x+(1/6))^2 =(1/9)⇔x=(1/3)−(1/6) or x=((−1)/3)−(1/6) so x=(1/6) or x=((−1)/2)

$$\mathrm{2}{sin}^{−\mathrm{1}} \left({x}\sqrt{\mathrm{6}}\right)+{sin}^{−\mathrm{1}} \left(\mathrm{4}{x}\right)=\frac{\pi}{\mathrm{2}} \\ $$$${sin}^{−\mathrm{1}} \left(\mathrm{4}{x}\right)=\frac{\pi}{\mathrm{2}}−\mathrm{2}{sin}^{−\mathrm{1}} \left({x}\sqrt{\mathrm{6}}\right) \\ $$$${sin}\left({sin}^{−\mathrm{1}} \left(\mathrm{4}{x}\right)\right)={sin}\left(\frac{\pi}{\mathrm{2}}−\mathrm{2}{sin}^{−\mathrm{1}} \left({x}\sqrt{\mathrm{6}}\right)\right) \\ $$$${we}\:{have}\:\:{sin}\left(\pi/\mathrm{2}−{a}\right)={cosa} \\ $$$${so}\::\:\:\mathrm{4}{x}={cos}\left(\mathrm{2}{sin}^{−\mathrm{1}} \left({x}\sqrt{\mathrm{6}}\right)=\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \left({sin}^{−\mathrm{1}} \left({x}\sqrt{\mathrm{6}}\right)\right)\right. \\ $$$$\mathrm{4}{x}=\mathrm{1}−\mathrm{2}×\left({x}\sqrt{\mathrm{6}}\right)^{\mathrm{2}} \Leftrightarrow\mathrm{12}{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{1}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}}{x}−\frac{\mathrm{1}}{\mathrm{12}}=\mathrm{0}\Leftrightarrow{x}^{\mathrm{2}} +\mathrm{2}.\frac{\mathrm{1}}{\mathrm{6}}{x}+\frac{\mathrm{1}}{\mathrm{36}}=\frac{\mathrm{1}}{\mathrm{36}}+\frac{\mathrm{1}}{\mathrm{12}}=\frac{\mathrm{1}}{\mathrm{9}} \\ $$$$\left({x}+\frac{\mathrm{1}}{\mathrm{6}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{9}}\Leftrightarrow{x}=\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{6}}\:\:{or}\:\:{x}=\frac{−\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{6}} \\ $$$${so}\:\:{x}=\frac{\mathrm{1}}{\mathrm{6}}\:\:{or}\:\:{x}=\frac{−\mathrm{1}}{\mathrm{2}} \\ $$

Commented by sma3l2996 last updated on 20/Sep/17

and −(1/2) it′s not solution, because (1/2)(√6)>1 or 4×(1/2)>1 so x=(1/6)

$${and}\:−\frac{\mathrm{1}}{\mathrm{2}}\:{it}'{s}\:{not}\:{solution},\:{because}\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{6}}>\mathrm{1}\:{or}\:\:\mathrm{4}×\frac{\mathrm{1}}{\mathrm{2}}>\mathrm{1} \\ $$$${so}\:\:{x}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$

Commented by NECx last updated on 20/Sep/17

$$\mathrm{i}'\mathrm{m}\:\mathrm{most}\:\mathrm{grateful}\:\mathrm{sir}. \\ $$

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