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Question-74945




Question Number 74945 by chess1 last updated on 04/Dec/19
Commented by chess1 last updated on 04/Dec/19
solve equation
$$\mathrm{solve}\:\mathrm{equation} \\ $$
Answered by MJS last updated on 04/Dec/19
solving for 0≤x<2π at first  cos x >0∧sin x >0 ⇒ 0<x<(π/2)  ((ln 24cos x)/(ln 24sin x))=(3/2)  ⇒  cos^2  x −24sin^3  x =0  ⇒  sin^3  x +(1/(24))sin^2  x −(1/(24))=0  (sin x −(1/3))(sin^2  x +(3/8)sin x +(1/8))=0  ⇒ sin x =(1/3) ⇒ x=arcsin (1/3)  generally x=2nπ+arcsin (1/3)
$$\mathrm{solving}\:\mathrm{for}\:\mathrm{0}\leqslant{x}<\mathrm{2}\pi\:\mathrm{at}\:\mathrm{first} \\ $$$$\mathrm{cos}\:{x}\:>\mathrm{0}\wedge\mathrm{sin}\:{x}\:>\mathrm{0}\:\Rightarrow\:\mathrm{0}<{x}<\frac{\pi}{\mathrm{2}} \\ $$$$\frac{\mathrm{ln}\:\mathrm{24cos}\:{x}}{\mathrm{ln}\:\mathrm{24sin}\:{x}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$$\mathrm{cos}^{\mathrm{2}} \:{x}\:−\mathrm{24sin}^{\mathrm{3}} \:{x}\:=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\mathrm{sin}^{\mathrm{3}} \:{x}\:+\frac{\mathrm{1}}{\mathrm{24}}\mathrm{sin}^{\mathrm{2}} \:{x}\:−\frac{\mathrm{1}}{\mathrm{24}}=\mathrm{0} \\ $$$$\left(\mathrm{sin}\:{x}\:−\frac{\mathrm{1}}{\mathrm{3}}\right)\left(\mathrm{sin}^{\mathrm{2}} \:{x}\:+\frac{\mathrm{3}}{\mathrm{8}}\mathrm{sin}\:{x}\:+\frac{\mathrm{1}}{\mathrm{8}}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{sin}\:{x}\:=\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\:{x}=\mathrm{arcsin}\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{generally}\:{x}=\mathrm{2}{n}\pi+\mathrm{arcsin}\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$

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