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Question Number 163822 by mkam last updated on 11/Jan/22
sin(4x) − sin(3x) = sin(2x) find x ?
$${sin}\left(\mathrm{4}{x}\right)\:−\:{sin}\left(\mathrm{3}{x}\right)\:=\:{sin}\left(\mathrm{2}{x}\right)\:{find}\:{x}\:? \\ $$
Answered by cortano1 last updated on 11/Jan/22
Commented by mr W last updated on 11/Jan/22
please recheck. i think answer is wrong.
$${please}\:{recheck}.\:{i}\:{think}\:{answer}\:{is} \\ $$$${wrong}. \\ $$
Commented by cortano1 last updated on 11/Jan/22
what wrong !
$${what}\:{wrong}\:! \\ $$
Commented by mr W last updated on 11/Jan/22
x=±cos^(−1) ((2−((10))^(1/3) −((100))^(1/3) )/6)+2kπ doesn′t fulfill the equation.
$${x}=\pm\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{2}−\sqrt[{\mathrm{3}}]{\mathrm{10}}−\sqrt[{\mathrm{3}}]{\mathrm{100}}}{\mathrm{6}}+\mathrm{2}{k}\pi\:{doesn}'{t} \\ $$$${fulfill}\:{the}\:{equation}. \\ $$
Answered by mr W last updated on 11/Jan/22
sin 4x−sin 2x=sin 3x sin (3x+x)−sin (3x−x)=sin 3x 2 cos 3x sin x=sin 3x 2(4 cos^3 x−3 cos x)sin x=sin x (3−4 sin^2 x) ⇒sin x=0 ⇒x=nπ or 8 cos^3 x−6 cos x=3−4 sin^2 x 8 cos^3 x−4 cos^2 x−6 cos x+1=0 cos^3 x−(1/2) cos^2 x−(3/4) cos x+(1/8)=0 let cos x=t+(1/6) t^3 −((5t)/6)−(1/(108))=0 t=((√(10))/3) sin (((2kπ)/3)−(1/3)sin^(−1) ((√(10))/(100))) cos x=(1/6)+((√(10))/3) sin (((2kπ)/3)−(1/3)sin^(−1) ((√(10))/(100))) such that −1≤cos x≤1, ⇒ k=0, 2 cos x=(1/6)−((√(10))/3) sin ((1/3)sin^(−1) ((√(10))/(100))) >0 cos x=(1/6)−((√(10))/3) sin ((π/3)−(1/3)sin^(−1) ((√(10))/(100))) <0 ⇒x=2nπ±cos^(−1) {(1/6)−((√(10))/3) sin ((1/3)sin^(−1) ((√(10))/(100)))} ⇒x=(2n+1)π±cos^(−1) {((√(10))/3) sin ((π/3)−(1/3)sin^(−1) ((√(10))/(100)))−(1/6)}
$$\mathrm{sin}\:\mathrm{4}{x}−\mathrm{sin}\:\mathrm{2}{x}=\mathrm{sin}\:\mathrm{3}{x} \\ $$$$\mathrm{sin}\:\left(\mathrm{3}{x}+{x}\right)−\mathrm{sin}\:\left(\mathrm{3}{x}−{x}\right)=\mathrm{sin}\:\mathrm{3}{x} \\ $$$$\mathrm{2}\:\mathrm{cos}\:\mathrm{3}{x}\:\mathrm{sin}\:{x}=\mathrm{sin}\:\mathrm{3}{x} \\ $$$$\mathrm{2}\left(\mathrm{4}\:\mathrm{cos}^{\mathrm{3}} \:{x}−\mathrm{3}\:\mathrm{cos}\:{x}\right)\mathrm{sin}\:{x}=\mathrm{sin}\:{x}\:\left(\mathrm{3}−\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:{x}\right) \\ $$$$\Rightarrow\mathrm{sin}\:{x}=\mathrm{0}\:\Rightarrow{x}={n}\pi \\ $$$${or} \\ $$$$\mathrm{8}\:\mathrm{cos}^{\mathrm{3}} \:{x}−\mathrm{6}\:\mathrm{cos}\:{x}=\mathrm{3}−\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:{x} \\ $$$$\mathrm{8}\:\mathrm{cos}^{\mathrm{3}} \:{x}−\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:{x}−\mathrm{6}\:\mathrm{cos}\:{x}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{cos}^{\mathrm{3}} \:{x}−\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{cos}^{\mathrm{2}} \:{x}−\frac{\mathrm{3}}{\mathrm{4}}\:\mathrm{cos}\:{x}+\frac{\mathrm{1}}{\mathrm{8}}=\mathrm{0} \\ $$$${let}\:\mathrm{cos}\:{x}={t}+\frac{\mathrm{1}}{\mathrm{6}} \\ $$$${t}^{\mathrm{3}} −\frac{\mathrm{5}{t}}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{108}}=\mathrm{0} \\ $$$${t}=\frac{\sqrt{\mathrm{10}}}{\mathrm{3}}\:\mathrm{sin}\:\left(\frac{\mathrm{2}{k}\pi}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{\mathrm{10}}}{\mathrm{100}}\right) \\ $$$$\mathrm{cos}\:{x}=\frac{\mathrm{1}}{\mathrm{6}}+\frac{\sqrt{\mathrm{10}}}{\mathrm{3}}\:\mathrm{sin}\:\left(\frac{\mathrm{2}{k}\pi}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{\mathrm{10}}}{\mathrm{100}}\right) \\ $$$${such}\:{that}\:−\mathrm{1}\leqslant\mathrm{cos}\:{x}\leqslant\mathrm{1},\:\Rightarrow\:{k}=\mathrm{0},\:\mathrm{2} \\ $$$$\mathrm{cos}\:{x}=\frac{\mathrm{1}}{\mathrm{6}}−\frac{\sqrt{\mathrm{10}}}{\mathrm{3}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{\mathrm{10}}}{\mathrm{100}}\right)\:>\mathrm{0} \\ $$$$\mathrm{cos}\:{x}=\frac{\mathrm{1}}{\mathrm{6}}−\frac{\sqrt{\mathrm{10}}}{\mathrm{3}}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{\mathrm{10}}}{\mathrm{100}}\right)\:<\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{2}{n}\pi\pm\mathrm{cos}^{−\mathrm{1}} \left\{\frac{\mathrm{1}}{\mathrm{6}}−\frac{\sqrt{\mathrm{10}}}{\mathrm{3}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{\mathrm{10}}}{\mathrm{100}}\right)\right\} \\ $$$$\Rightarrow{x}=\left(\mathrm{2}{n}+\mathrm{1}\right)\pi\pm\mathrm{cos}^{−\mathrm{1}} \left\{\frac{\sqrt{\mathrm{10}}}{\mathrm{3}}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{\mathrm{10}}}{\mathrm{100}}\right)−\frac{\mathrm{1}}{\mathrm{6}}\right\} \\ $$
Commented by peter frank last updated on 11/Jan/22
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$

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