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Question-72291




Question Number 72291 by aliesam last updated on 27/Oct/19
Answered by mr W last updated on 27/Oct/19
(5m sin (3x))^2 =(4 sin x)^2   5m sin (3x)=±4 sin x  5m sin x (3−4 sin^2  x)=±4 sin x  ⇒sin x=0 ⇒x=kπ  ⇒5m(3−4 sin^2  x)=±4  ⇒sin^2  x=(1/4)(3±(4/(5m)))  ⇒sin x=±(1/2)(√(3±(4/(5m))))  ⇒x=kπ±sin^(−1) (1/2)(√(3±(4/(5m))))  with 0≤3±(4/(5m))≤4  ⇒m≥(4/5)  ⇒m≥(4/(15))
$$\left(\mathrm{5}{m}\:\mathrm{sin}\:\left(\mathrm{3}{x}\right)\right)^{\mathrm{2}} =\left(\mathrm{4}\:\mathrm{sin}\:{x}\right)^{\mathrm{2}} \\ $$$$\mathrm{5}{m}\:\mathrm{sin}\:\left(\mathrm{3}{x}\right)=\pm\mathrm{4}\:\mathrm{sin}\:{x} \\ $$$$\mathrm{5}{m}\:\mathrm{sin}\:{x}\:\left(\mathrm{3}−\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:{x}\right)=\pm\mathrm{4}\:\mathrm{sin}\:{x} \\ $$$$\Rightarrow\mathrm{sin}\:{x}=\mathrm{0}\:\Rightarrow{x}={k}\pi \\ $$$$\Rightarrow\mathrm{5}{m}\left(\mathrm{3}−\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:{x}\right)=\pm\mathrm{4} \\ $$$$\Rightarrow\mathrm{sin}^{\mathrm{2}} \:{x}=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{3}\pm\frac{\mathrm{4}}{\mathrm{5}{m}}\right) \\ $$$$\Rightarrow\mathrm{sin}\:{x}=\pm\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}\pm\frac{\mathrm{4}}{\mathrm{5}{m}}} \\ $$$$\Rightarrow{x}={k}\pi\pm\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}\pm\frac{\mathrm{4}}{\mathrm{5}{m}}} \\ $$$${with}\:\mathrm{0}\leqslant\mathrm{3}\pm\frac{\mathrm{4}}{\mathrm{5}{m}}\leqslant\mathrm{4} \\ $$$$\Rightarrow{m}\geqslant\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\Rightarrow{m}\geqslant\frac{\mathrm{4}}{\mathrm{15}} \\ $$

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