Question and Answers Forum

All Questions      Topic List

Logarithms Questions

Previous in All Question      Next in All Question      

Previous in Logarithms      Next in Logarithms      

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)

$$\mathrm{solving}\:\mathrm{for}\:\mathrm{0}\leqslant{x}<\mathrm{2}\pi\:\mathrm{at}\:\mathrm{first} \\ $$$$\mathrm{cos}\:{x}\:>\mathrm{0}\wedge\mathrm{sin}\:{x}\:>\mathrm{0}\:\Rightarrow\:\mathrm{0}<{x}<\frac{\pi}{\mathrm{2}} \\ $$$$\frac{\mathrm{ln}\:\mathrm{24cos}\:{x}}{\mathrm{ln}\:\mathrm{24sin}\:{x}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$$\mathrm{cos}^{\mathrm{2}} \:{x}\:−\mathrm{24sin}^{\mathrm{3}} \:{x}\:=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\mathrm{sin}^{\mathrm{3}} \:{x}\:+\frac{\mathrm{1}}{\mathrm{24}}\mathrm{sin}^{\mathrm{2}} \:{x}\:−\frac{\mathrm{1}}{\mathrm{24}}=\mathrm{0} \\ $$$$\left(\mathrm{sin}\:{x}\:−\frac{\mathrm{1}}{\mathrm{3}}\right)\left(\mathrm{sin}^{\mathrm{2}} \:{x}\:+\frac{\mathrm{3}}{\mathrm{8}}\mathrm{sin}\:{x}\:+\frac{\mathrm{1}}{\mathrm{8}}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{sin}\:{x}\:=\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\:{x}=\mathrm{arcsin}\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{generally}\:{x}=\mathrm{2}{n}\pi+\mathrm{arcsin}\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com