Question Number 211657 by hardmath last updated on 15/Sep/24
$$\mathrm{sin}\left(\mathrm{9x}\right)\:=\:\mathrm{sin}\left(\mathrm{5x}\right)\:+\:\mathrm{sin}\left(\mathrm{3x}\right) \\ $$$$\mathrm{find}:\:\:\mathrm{x}\:=\:? \\ $$
Answered by Frix last updated on 15/Sep/24
![sin ((2n+1)x) =sin x (1+2Σ_(k=1) ^n cos (2kx)) ⇒ sin 9x −sin 5x −sin 3x =0 ⇔ 2(cos 8x +cos 6x −cos 2x −(1/2))sin x =0 Let 0≤x<2π ★ x=0∨x=π cos 8x +cos 6x −cos 2x −(1/2)=0 This has a period of π and it′s symmetric to ((nπ)/2) ⇒ We must look for solutions x∈[0; (π/2)] These should be at (((2k+1)π)/(30)) Testing leads to ★ x=(π/(30)), ((7π)/(30)), ((11π)/(30)), ((13π)/(30)) [0; (π/2)] ⇒ Due to symmetry ★ x=((17π)/(30)), ((19π)/(39)), ((23π)/(30)), ((29π)/(30)) [(π/2); π] ★ All these +π [π; 2π) We get x=(π/(30)){0, 1, 7, 11, 13, 17, 19, 23, 29}+nπ](https://www.tinkutara.com/question/Q211670.png)
$$\mathrm{sin}\:\left(\left(\mathrm{2}{n}+\mathrm{1}\right){x}\right)\:=\mathrm{sin}\:{x}\:\left(\mathrm{1}+\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\mathrm{cos}\:\left(\mathrm{2}{kx}\right)\right) \\ $$$$\Rightarrow \\ $$$$\mathrm{sin}\:\mathrm{9}{x}\:−\mathrm{sin}\:\mathrm{5}{x}\:−\mathrm{sin}\:\mathrm{3}{x}\:=\mathrm{0} \\ $$$$\Leftrightarrow \\ $$$$\mathrm{2}\left(\mathrm{cos}\:\mathrm{8}{x}\:+\mathrm{cos}\:\mathrm{6}{x}\:−\mathrm{cos}\:\mathrm{2}{x}\:−\frac{\mathrm{1}}{\mathrm{2}}\right)\mathrm{sin}\:{x}\:=\mathrm{0} \\ $$$$\mathrm{Let}\:\mathrm{0}\leqslant{x}<\mathrm{2}\pi \\ $$$$\bigstar\:{x}=\mathrm{0}\vee{x}=\pi \\ $$$$\mathrm{cos}\:\mathrm{8}{x}\:+\mathrm{cos}\:\mathrm{6}{x}\:−\mathrm{cos}\:\mathrm{2}{x}\:−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$$\mathrm{This}\:\mathrm{has}\:\mathrm{a}\:\mathrm{period}\:\mathrm{of}\:\pi\:\mathrm{and}\:\mathrm{it}'\mathrm{s}\:\mathrm{symmetric} \\ $$$$\mathrm{to}\:\frac{{n}\pi}{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{We}\:\mathrm{must}\:\mathrm{look}\:\mathrm{for}\:\mathrm{solutions}\:{x}\in\left[\mathrm{0};\:\frac{\pi}{\mathrm{2}}\right] \\ $$$$\mathrm{These}\:\mathrm{should}\:\mathrm{be}\:\mathrm{at}\:\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{30}} \\ $$$$\mathrm{Testing}\:\mathrm{leads}\:\mathrm{to} \\ $$$$\bigstar\:{x}=\frac{\pi}{\mathrm{30}},\:\frac{\mathrm{7}\pi}{\mathrm{30}},\:\frac{\mathrm{11}\pi}{\mathrm{30}},\:\frac{\mathrm{13}\pi}{\mathrm{30}}\:\:\:\:\:\left[\mathrm{0};\:\frac{\pi}{\mathrm{2}}\right] \\ $$$$\Rightarrow\:\mathrm{Due}\:\mathrm{to}\:\mathrm{symmetry} \\ $$$$\bigstar\:{x}=\frac{\mathrm{17}\pi}{\mathrm{30}},\:\frac{\mathrm{19}\pi}{\mathrm{39}},\:\frac{\mathrm{23}\pi}{\mathrm{30}},\:\frac{\mathrm{29}\pi}{\mathrm{30}}\:\:\:\:\:\left[\frac{\pi}{\mathrm{2}};\:\pi\right] \\ $$$$\bigstar\:\mathrm{All}\:\mathrm{these}\:+\pi\:\:\:\:\:\left[\pi;\:\mathrm{2}\pi\right) \\ $$$$ \\ $$$$\mathrm{We}\:\mathrm{get} \\ $$$${x}=\frac{\pi}{\mathrm{30}}\left\{\mathrm{0},\:\mathrm{1},\:\mathrm{7},\:\mathrm{11},\:\mathrm{13},\:\mathrm{17},\:\mathrm{19},\:\mathrm{23},\:\mathrm{29}\right\}+{n}\pi \\ $$