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Question Number 40285 by MJS last updated on 18/Jul/18

solve for 1 period  sin (α/2) =sin 2α  sin (β/3) =sin 3β  cos (γ/2) =cos 2γ  cos (δ/3) =cos 3δ  tan (ε/2) =tan 2ε  tan (ζ/3) =tan 3ζ

$$\mathrm{solve}\:\mathrm{for}\:\mathrm{1}\:\mathrm{period} \\ $$$$\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}\:=\mathrm{sin}\:\mathrm{2}\alpha \\ $$$$\mathrm{sin}\:\frac{\beta}{\mathrm{3}}\:=\mathrm{sin}\:\mathrm{3}\beta \\ $$$$\mathrm{cos}\:\frac{\gamma}{\mathrm{2}}\:=\mathrm{cos}\:\mathrm{2}\gamma \\ $$$$\mathrm{cos}\:\frac{\delta}{\mathrm{3}}\:=\mathrm{cos}\:\mathrm{3}\delta \\ $$$$\mathrm{tan}\:\frac{\epsilon}{\mathrm{2}}\:=\mathrm{tan}\:\mathrm{2}\epsilon \\ $$$$\mathrm{tan}\:\frac{\zeta}{\mathrm{3}}\:=\mathrm{tan}\:\mathrm{3}\zeta \\ $$

Answered by MrW3 last updated on 19/Jul/18

sin (α/2) =sin 2α  ⇒2α+2mπ=(α/2)+2nπ⇒α=((4kπ)/3)  or  ⇒2α+2mπ=(2n+1)π−(α/2)⇒α=((2(2k+1)π)/5)    ⇒solution is:  α=((4kπ)/3) and α=((2(2k+1)π)/5) with k=0,±1,±2,...    one period: 0≤α≤4π  ((4kπ)/3)≤4π⇒k≤3  ((2(2k+1)π)/5)≤4π⇒k≤4.5  ⇒α=0,((2π)/5),((6π)/5),((4π)/3),2π,((8π)/3),((14π)/5),((18π)/5),4π

$$\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}\:=\mathrm{sin}\:\mathrm{2}\alpha \\ $$$$\Rightarrow\mathrm{2}\alpha+\mathrm{2}{m}\pi=\frac{\alpha}{\mathrm{2}}+\mathrm{2}{n}\pi\Rightarrow\alpha=\frac{\mathrm{4}{k}\pi}{\mathrm{3}} \\ $$$${or} \\ $$$$\Rightarrow\mathrm{2}\alpha+\mathrm{2}{m}\pi=\left(\mathrm{2}{n}+\mathrm{1}\right)\pi−\frac{\alpha}{\mathrm{2}}\Rightarrow\alpha=\frac{\mathrm{2}\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{5}} \\ $$$$ \\ $$$$\Rightarrow{solution}\:{is}: \\ $$$$\alpha=\frac{\mathrm{4}{k}\pi}{\mathrm{3}}\:{and}\:\alpha=\frac{\mathrm{2}\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{5}}\:{with}\:{k}=\mathrm{0},\pm\mathrm{1},\pm\mathrm{2},... \\ $$$$ \\ $$$${one}\:{period}:\:\mathrm{0}\leqslant\alpha\leqslant\mathrm{4}\pi \\ $$$$\frac{\mathrm{4}{k}\pi}{\mathrm{3}}\leqslant\mathrm{4}\pi\Rightarrow{k}\leqslant\mathrm{3} \\ $$$$\frac{\mathrm{2}\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{5}}\leqslant\mathrm{4}\pi\Rightarrow{k}\leqslant\mathrm{4}.\mathrm{5} \\ $$$$\Rightarrow\alpha=\mathrm{0},\frac{\mathrm{2}\pi}{\mathrm{5}},\frac{\mathrm{6}\pi}{\mathrm{5}},\frac{\mathrm{4}\pi}{\mathrm{3}},\mathrm{2}\pi,\frac{\mathrm{8}\pi}{\mathrm{3}},\frac{\mathrm{14}\pi}{\mathrm{5}},\frac{\mathrm{18}\pi}{\mathrm{5}},\mathrm{4}\pi \\ $$

Answered by MrW3 last updated on 19/Jul/18

tan (ε/2) =tan 2ε  (ε/2)+mπ=2ε+nπ ⇒ε=((2kπ)/3)  ⇒solution is:  ε=((2kπ)/3)

$$\mathrm{tan}\:\frac{\epsilon}{\mathrm{2}}\:=\mathrm{tan}\:\mathrm{2}\epsilon \\ $$$$\frac{\epsilon}{\mathrm{2}}+{m}\pi=\mathrm{2}\epsilon+{n}\pi\:\Rightarrow\epsilon=\frac{\mathrm{2}{k}\pi}{\mathrm{3}} \\ $$$$\Rightarrow{solution}\:{is}: \\ $$$$\epsilon=\frac{\mathrm{2}{k}\pi}{\mathrm{3}} \\ $$

Answered by MrW3 last updated on 19/Jul/18

cos (γ/2) =cos 2γ  (γ/2)+2mπ=2nπ±2γ⇒ { ((γ=((4kπ)/3))),((γ=((4kπ)/5))) :}  ⇒solution is:  γ=((4kπ)/3) and γ=((4kπ)/5)

$$\mathrm{cos}\:\frac{\gamma}{\mathrm{2}}\:=\mathrm{cos}\:\mathrm{2}\gamma \\ $$$$\frac{\gamma}{\mathrm{2}}+\mathrm{2}{m}\pi=\mathrm{2}{n}\pi\pm\mathrm{2}\gamma\Rightarrow\begin{cases}{\gamma=\frac{\mathrm{4}{k}\pi}{\mathrm{3}}}\\{\gamma=\frac{\mathrm{4}{k}\pi}{\mathrm{5}}}\end{cases} \\ $$$$\Rightarrow{solution}\:{is}: \\ $$$$\gamma=\frac{\mathrm{4}{k}\pi}{\mathrm{3}}\:{and}\:\gamma=\frac{\mathrm{4}{k}\pi}{\mathrm{5}} \\ $$

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