Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 157044 by MathSh last updated on 18/Oct/21

Solve for real numbers:  (sin2x + 4cos^2 x + 1)(cos5x - cosx)<0

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$ $$\left(\mathrm{sin2}\boldsymbol{\mathrm{x}}\:+\:\mathrm{4cos}^{\mathrm{2}} \boldsymbol{\mathrm{x}}\:+\:\mathrm{1}\right)\left(\mathrm{cos5}\boldsymbol{\mathrm{x}}\:-\:\mathrm{cos}\boldsymbol{\mathrm{x}}\right)<\mathrm{0} \\ $$

Answered by mindispower last updated on 19/Oct/21

1+sin(2x)=(sin(x)+cos(x))^2   ⇔((sin(x)+cos(x))^2 +4cos^2 (x))≥0  cos(5x)−cos(x)=−2sin(3x)sin(2x)  ⇔sin(2x)sin(3x)>0  not hard start her

$$\mathrm{1}+{sin}\left(\mathrm{2}{x}\right)=\left({sin}\left({x}\right)+{cos}\left({x}\right)\right)^{\mathrm{2}} \\ $$ $$\Leftrightarrow\left(\left({sin}\left({x}\right)+{cos}\left({x}\right)\right)^{\mathrm{2}} +\mathrm{4}{cos}^{\mathrm{2}} \left({x}\right)\right)\geqslant\mathrm{0} \\ $$ $${cos}\left(\mathrm{5}{x}\right)−{cos}\left({x}\right)=−\mathrm{2}\boldsymbol{{sin}}\left(\mathrm{3}\boldsymbol{{x}}\right)\boldsymbol{{sin}}\left(\mathrm{2}\boldsymbol{{x}}\right) \\ $$ $$\Leftrightarrow{sin}\left(\mathrm{2}{x}\right){sin}\left(\mathrm{3}{x}\right)>\mathrm{0} \\ $$ $${not}\:{hard}\:{start}\:{her}\: \\ $$

Commented byMathSh last updated on 19/Oct/21

Very nice dear Ser thank you

$$\mathrm{Very}\:\mathrm{nice}\:\mathrm{dear}\:\mathrm{Ser}\:\mathrm{thank}\:\mathrm{you} \\ $$

Commented bymindispower last updated on 19/Oct/21

withe pleasur

$${withe}\:{pleasur} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com