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

cos^(−1) (2x)+ cos^(−1) (x)=(π/6) find x

$$\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)+\:\mathrm{cos}^{−\mathrm{1}} \left({x}\right)=\frac{\pi}{\mathrm{6}} \\ $$$${find}\:{x} \\ $$

Answered by bramlex last updated on 30/Jul/20

cos^(−1) (2x)=(π/6)−cos^(−1) (x) cos (cos^(−1) (2x))= cos ((π/6)−cos^(−1) (x)) 2x = ((√3)/2) x+(1/2)sin (cos^(−1) (x)) 4x=(√3) x+sin (cos^(−1) (x)) x(4−(√3))=sin (cos^(−1) (x)) x(4−(√3)) = (√(1−cos^2 (cos^(−1) (x)))) x(4−(√3)) = (√(1−x^2 )) x^2 (4−(√3))^2 =1−x^2 x^2 (19−8(√3)+1) = 1 x^2 = (1/(20−8(√3))) = (1/(4(5−2(√3)))) x = ±(1/(2(√(5−2(√3))))) .▲

$$\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)=\frac{\pi}{\mathrm{6}}−\mathrm{cos}^{−\mathrm{1}} \left({x}\right) \\ $$$$\mathrm{cos}\:\left(\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)\right)=\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{6}}−\mathrm{cos}^{−\mathrm{1}} \left({x}\right)\right) \\ $$$$\mathrm{2}{x}\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{x}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\left(\mathrm{cos}^{−\mathrm{1}} \left({x}\right)\right) \\ $$$$\mathrm{4}{x}=\sqrt{\mathrm{3}}\:{x}+\mathrm{sin}\:\left(\mathrm{cos}^{−\mathrm{1}} \left({x}\right)\right) \\ $$$${x}\left(\mathrm{4}−\sqrt{\mathrm{3}}\right)=\mathrm{sin}\:\left(\mathrm{cos}^{−\mathrm{1}} \left({x}\right)\right) \\ $$$${x}\left(\mathrm{4}−\sqrt{\mathrm{3}}\right)\:=\:\sqrt{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{cos}^{−\mathrm{1}} \left({x}\right)\right)} \\ $$$${x}\left(\mathrm{4}−\sqrt{\mathrm{3}}\right)\:=\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$${x}^{\mathrm{2}} \left(\mathrm{4}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} =\mathrm{1}−{x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} \left(\mathrm{19}−\mathrm{8}\sqrt{\mathrm{3}}+\mathrm{1}\right)\:=\:\mathrm{1} \\ $$$${x}^{\mathrm{2}} =\:\frac{\mathrm{1}}{\mathrm{20}−\mathrm{8}\sqrt{\mathrm{3}}}\:=\:\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{5}−\mathrm{2}\sqrt{\mathrm{3}}\right)} \\ $$$${x}\:=\:\pm\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{5}−\mathrm{2}\sqrt{\mathrm{3}}}}\:.\blacktriangle \\ $$

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

⇒cos(arcos(2x)+cosx) =((√3)/2) ⇒ 2x .x −sin(arcos(2x))sin(cosx) =((√3)/2) ⇒ 2x^2 −(√(1−(2x)^2 ))×(√(1−x^2 ))=((√3)/2) ⇒ (2x^2 −((√3)/2)) =(√(1−4x^2 ))×(√(1−x^2 )) ⇒ (2x^2 −((√3)/2))^2 =(1−4x^2 )(1−x^2 ) ⇒ 4x^4 −2(√3)x^2 +(3/4) =1−x^2 −4x^2 +4x^4 ⇒ −2(√3)x^2 +(3/4) =1−5x^2 ⇒−8(√3)x^2 +3 =4−20x^2 ⇒ (20−8(√3))x^2 =−3 ⇒(8(√3)−20)x^2 =3 ⇒x^2 =(3/(8(√3)−20)) ⇒ x =+^− (√(3/(8(√3)−20)))

$$\Rightarrow\mathrm{cos}\left(\mathrm{arcos}\left(\mathrm{2x}\right)+\mathrm{cosx}\right)\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{2x}\:.\mathrm{x}\:−\mathrm{sin}\left(\mathrm{arcos}\left(\mathrm{2x}\right)\right)\mathrm{sin}\left(\mathrm{cosx}\right)\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{2x}^{\mathrm{2}} −\sqrt{\mathrm{1}−\left(\mathrm{2x}\right)^{\mathrm{2}} }×\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\Rightarrow \\ $$$$\left(\mathrm{2x}^{\mathrm{2}} −\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\:=\sqrt{\mathrm{1}−\mathrm{4x}^{\mathrm{2}} }×\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\left(\mathrm{2x}^{\mathrm{2}} −\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} \:=\left(\mathrm{1}−\mathrm{4x}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$$\mathrm{4x}^{\mathrm{4}} −\mathrm{2}\sqrt{\mathrm{3}}\mathrm{x}^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}\:=\mathrm{1}−\mathrm{x}^{\mathrm{2}} −\mathrm{4x}^{\mathrm{2}} +\mathrm{4x}^{\mathrm{4}} \:\Rightarrow \\ $$$$−\mathrm{2}\sqrt{\mathrm{3}}\mathrm{x}^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}\:=\mathrm{1}−\mathrm{5x}^{\mathrm{2}} \:\Rightarrow−\mathrm{8}\sqrt{\mathrm{3}}\mathrm{x}^{\mathrm{2}} +\mathrm{3}\:=\mathrm{4}−\mathrm{20x}^{\mathrm{2}} \:\Rightarrow \\ $$$$\left(\mathrm{20}−\mathrm{8}\sqrt{\mathrm{3}}\right)\mathrm{x}^{\mathrm{2}} \:=−\mathrm{3}\:\Rightarrow\left(\mathrm{8}\sqrt{\mathrm{3}}−\mathrm{20}\right)\mathrm{x}^{\mathrm{2}} \:=\mathrm{3}\:\Rightarrow\mathrm{x}^{\mathrm{2}} \:=\frac{\mathrm{3}}{\mathrm{8}\sqrt{\mathrm{3}}−\mathrm{20}}\:\Rightarrow \\ $$$$\mathrm{x}\:=\overset{−} {+}\sqrt{\frac{\mathrm{3}}{\mathrm{8}\sqrt{\mathrm{3}}−\mathrm{20}}} \\ $$

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