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Question-108016




Question Number 108016 by mathdave last updated on 13/Aug/20
Answered by mr W last updated on 13/Aug/20
θ=cos^(−1) x ⇒ 0≤θ≤π ⇒0≤3θ≤3π  cos θ=x  φ=sin^(−1) 2x ⇒−(π/2)≤φ≤(π/2) ⇒−π≤2φ≤π  sin φ=2x=2 cos θ  2φ=3θ ⇒0≤3θ≤π ⇒0≤θ≤(π/3) ⇒(1/2)≤cos θ≤1  cos (2φ)=cos (3θ)  1−2 sin^2  φ=4 cos^3  θ−3 cos θ  1−8 cos^2  θ=4 cos^3  θ−3 cos θ  ⇒4 cos^3  θ+8 cos^2  θ−3 cos θ−1=0  ⇒(2 cos θ−1)(2 cos^2  θ+5 cos θ+1)=0  ⇒cos θ=(1/2)  ⇒cos θ=((−5±(√(17)))/2)  since (1/2)≤cos θ≤1  ⇒x=cos θ=(1/2) ⇒only one solution
$$\theta=\mathrm{cos}^{−\mathrm{1}} {x}\:\Rightarrow\:\mathrm{0}\leqslant\theta\leqslant\pi\:\Rightarrow\mathrm{0}\leqslant\mathrm{3}\theta\leqslant\mathrm{3}\pi \\ $$$$\mathrm{cos}\:\theta={x} \\ $$$$\phi=\mathrm{sin}^{−\mathrm{1}} \mathrm{2}{x}\:\Rightarrow−\frac{\pi}{\mathrm{2}}\leqslant\phi\leqslant\frac{\pi}{\mathrm{2}}\:\Rightarrow−\pi\leqslant\mathrm{2}\phi\leqslant\pi \\ $$$$\mathrm{sin}\:\phi=\mathrm{2}{x}=\mathrm{2}\:\mathrm{cos}\:\theta \\ $$$$\mathrm{2}\phi=\mathrm{3}\theta\:\Rightarrow\mathrm{0}\leqslant\mathrm{3}\theta\leqslant\pi\:\Rightarrow\mathrm{0}\leqslant\theta\leqslant\frac{\pi}{\mathrm{3}}\:\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\leqslant\mathrm{cos}\:\theta\leqslant\mathrm{1} \\ $$$$\mathrm{cos}\:\left(\mathrm{2}\phi\right)=\mathrm{cos}\:\left(\mathrm{3}\theta\right) \\ $$$$\mathrm{1}−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\phi=\mathrm{4}\:\mathrm{cos}^{\mathrm{3}} \:\theta−\mathrm{3}\:\mathrm{cos}\:\theta \\ $$$$\mathrm{1}−\mathrm{8}\:\mathrm{cos}^{\mathrm{2}} \:\theta=\mathrm{4}\:\mathrm{cos}^{\mathrm{3}} \:\theta−\mathrm{3}\:\mathrm{cos}\:\theta \\ $$$$\Rightarrow\mathrm{4}\:\mathrm{cos}^{\mathrm{3}} \:\theta+\mathrm{8}\:\mathrm{cos}^{\mathrm{2}} \:\theta−\mathrm{3}\:\mathrm{cos}\:\theta−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{2}\:\mathrm{cos}\:\theta−\mathrm{1}\right)\left(\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:\theta+\mathrm{5}\:\mathrm{cos}\:\theta+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{cos}\:\theta=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{cos}\:\theta=\frac{−\mathrm{5}\pm\sqrt{\mathrm{17}}}{\mathrm{2}} \\ $$$${since}\:\frac{\mathrm{1}}{\mathrm{2}}\leqslant\mathrm{cos}\:\theta\leqslant\mathrm{1} \\ $$$$\Rightarrow{x}=\mathrm{cos}\:\theta=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{only}\:{one}\:{solution} \\ $$

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