Question Number 186236 by ajfour last updated on 02/Feb/23
Answered by a.lgnaoui last updated on 02/Feb/23
$$\mathrm{1} \\ $$
Commented by ajfour last updated on 02/Feb/23
$${Its}\:{quite}\:{simple}\:{x}_{{min}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$
Commented by mr W last updated on 02/Feb/23
$${when}\:\alpha\rightarrow\mathrm{0},\:{x}\rightarrow\mathrm{0}. \\ $$$${that}\:{means}\:{there}\:{is}\:{no}\:{minimum}\:{for} \\ $$$${x}. \\ $$
Commented by mr W last updated on 02/Feb/23
Commented by mr W last updated on 03/Feb/23
$$\frac{{x}}{\mathrm{1}}=\frac{\mathrm{sin}\:\alpha}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\alpha+\alpha\right)} \\ $$$${x}=\frac{\mathrm{sin}\:\alpha}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\mathrm{2}\alpha\right)}\:\underset{\alpha\rightarrow\mathrm{0}} {\rightarrow}\mathrm{0} \\ $$$${x}=\frac{\mathrm{sin}\:\alpha}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\mathrm{2}\alpha\right)}\:\underset{\alpha\rightarrow\frac{\pi}{\mathrm{3}}} {\rightarrow}\infty \\ $$$$\Rightarrow{x}\:{has}\:{no}\:{minimum}\:{and}\:{maximum}. \\ $$