Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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

$$\theta=\mathrm{cos}^{−\mathrm{1}} {x}\:\Rightarrow\:\mathrm{0}\leqslant\theta\leqslant\pi\:\Rightarrow\mathrm{0}\leqslant\mathrm{3}\theta\leqslant\mathrm{3}\pi \\ $$$$\mathrm{cos}\:\theta={x} \\ $$$$\phi=\mathrm{sin}^{−\mathrm{1}} \mathrm{2}{x}\:\Rightarrow−\frac{\pi}{\mathrm{2}}\leqslant\phi\leqslant\frac{\pi}{\mathrm{2}}\:\Rightarrow−\pi\leqslant\mathrm{2}\phi\leqslant\pi \\ $$$$\mathrm{sin}\:\phi=\mathrm{2}{x}=\mathrm{2}\:\mathrm{cos}\:\theta \\ $$$$\mathrm{2}\phi=\mathrm{3}\theta\:\Rightarrow\mathrm{0}\leqslant\mathrm{3}\theta\leqslant\pi\:\Rightarrow\mathrm{0}\leqslant\theta\leqslant\frac{\pi}{\mathrm{3}}\:\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\leqslant\mathrm{cos}\:\theta\leqslant\mathrm{1} \\ $$$$\mathrm{cos}\:\left(\mathrm{2}\phi\right)=\mathrm{cos}\:\left(\mathrm{3}\theta\right) \\ $$$$\mathrm{1}−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\phi=\mathrm{4}\:\mathrm{cos}^{\mathrm{3}} \:\theta−\mathrm{3}\:\mathrm{cos}\:\theta \\ $$$$\mathrm{1}−\mathrm{8}\:\mathrm{cos}^{\mathrm{2}} \:\theta=\mathrm{4}\:\mathrm{cos}^{\mathrm{3}} \:\theta−\mathrm{3}\:\mathrm{cos}\:\theta \\ $$$$\Rightarrow\mathrm{4}\:\mathrm{cos}^{\mathrm{3}} \:\theta+\mathrm{8}\:\mathrm{cos}^{\mathrm{2}} \:\theta−\mathrm{3}\:\mathrm{cos}\:\theta−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{2}\:\mathrm{cos}\:\theta−\mathrm{1}\right)\left(\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:\theta+\mathrm{5}\:\mathrm{cos}\:\theta+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{cos}\:\theta=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{cos}\:\theta=\frac{−\mathrm{5}\pm\sqrt{\mathrm{17}}}{\mathrm{2}} \\ $$$${since}\:\frac{\mathrm{1}}{\mathrm{2}}\leqslant\mathrm{cos}\:\theta\leqslant\mathrm{1} \\ $$$$\Rightarrow{x}=\mathrm{cos}\:\theta=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{only}\:{one}\:{solution} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com