Question Number 159325 by mnjuly1970 last updated on 15/Nov/21
$$ \\ $$$$\:\:\:\:\:\:#\:\mathrm{T}{rigonometry}# \\ $$$$\:\:\:\:\:\:\:{solve}\:\left(\:\:\:\mathscr{E}{quation}\right) \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:{sin}\left(\frac{{x}}{\mathrm{2}}\:\right)\:−\:\mathrm{2}{sin}\:\left(\frac{{x}}{\mathrm{3}}\:\right)=\:\mathrm{0}\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$$$ \\ $$
Commented by Ar Brandon last updated on 15/Nov/21
$${x}=\mathrm{12}\pi{n}\pm\mathrm{12arctan}\left(\sqrt{\frac{\mathrm{5}}{\mathrm{7}}+\frac{\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{7}}}\right) \\ $$
Answered by mr W last updated on 15/Nov/21
$$\mathrm{sin}\:\left(\frac{\mathrm{3}{x}}{\mathrm{6}}\right)−\mathrm{2sin}\:\left(\frac{\mathrm{2}{x}}{\mathrm{6}}\right)=\mathrm{0} \\ $$$${let}\:{t}=\frac{{x}}{\mathrm{6}} \\ $$$$\mathrm{sin}\:\left(\mathrm{3}{t}\right)−\mathrm{2sin}\:\left(\mathrm{2}{t}\right)=\mathrm{0} \\ $$$$\mathrm{sin}\:{t}\:\left(\mathrm{3}−\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:{t}\right)−\mathrm{4sin}\:{t}\:\mathrm{cos}\:{t}=\mathrm{0} \\ $$$$\mathrm{sin}\:{t}\:\left(\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:{t}−\mathrm{4}\:\mathrm{cos}\:{t}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{sin}\:{t}=\mathrm{0}\:\Rightarrow{t}={k}\pi\:\Rightarrow{x}=\mathrm{6}{k}\pi \\ $$$$\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:{t}−\mathrm{4}\:\mathrm{cos}\:{t}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{cos}\:{t}=\frac{\mathrm{1}\pm\sqrt{\mathrm{2}}}{\mathrm{2}}=−\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{2}}\:\:\left(+\:{rejected}\right) \\ $$$$\Rightarrow{t}=\left(\mathrm{2}{k}+\mathrm{1}\right)\pi\pm\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{x}=\mathrm{6}\left(\mathrm{2}{k}+\mathrm{1}\right)\pi\pm\mathrm{6}\:\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{2}} \\ $$
Commented by mnjuly1970 last updated on 16/Nov/21
$${very}\:{nice}\:{sir}\:{W} \\ $$