Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 112173 by bobhans last updated on 06/Sep/20

find the solution of equation (√(cos 2x−sin^3 x+3)) = sin x

$$\:\mathrm{find}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{equation}\: \\ $$$$\sqrt{\mathrm{cos}\:\mathrm{2x}−\mathrm{sin}\:^{\mathrm{3}} \mathrm{x}+\mathrm{3}}\:=\:\mathrm{sin}\:\mathrm{x}\: \\ $$

Answered by john santu last updated on 06/Sep/20

since sin x ≥ 0 , then cos 2x−sin^3 x+3 = sin^2 x 1−2sin^2 x−sin^3 x+3 = sin^2 x sin^3 x+3sin^2 x−4 =0 (sin x−1)(sin^2 +4sin x+4)=0 (sin x−1)(sin x+2)^2 = 0 sin x = 1 →x = (π/2) + k.2π ; k∈Z

$${since}\:\mathrm{sin}\:{x}\:\geqslant\:\mathrm{0}\:,\:{then}\: \\ $$$$\mathrm{cos}\:\mathrm{2}{x}−\mathrm{sin}\:^{\mathrm{3}} {x}+\mathrm{3}\:=\:\mathrm{sin}\:^{\mathrm{2}} {x} \\ $$$$\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} {x}−\mathrm{sin}\:^{\mathrm{3}} {x}+\mathrm{3}\:=\:\mathrm{sin}\:^{\mathrm{2}} {x} \\ $$$$\mathrm{sin}\:^{\mathrm{3}} {x}+\mathrm{3sin}\:^{\mathrm{2}} {x}−\mathrm{4}\:=\mathrm{0} \\ $$$$\left(\mathrm{sin}\:{x}−\mathrm{1}\right)\left(\mathrm{sin}\:^{\mathrm{2}} +\mathrm{4sin}\:{x}+\mathrm{4}\right)=\mathrm{0} \\ $$$$\left(\mathrm{sin}\:{x}−\mathrm{1}\right)\left(\mathrm{sin}\:{x}+\mathrm{2}\right)^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\mathrm{sin}\:{x}\:=\:\mathrm{1}\:\rightarrow{x}\:=\:\frac{\pi}{\mathrm{2}}\:+\:{k}.\mathrm{2}\pi\:;\:{k}\in\mathbb{Z} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com