1-2sin-4x-lt-cos-2-4x-solve-

Question Number 57357 by ANTARES VY last updated on 03/Apr/19

$$\mathrm{1}−\mathrm{2}\boldsymbol{\mathrm{sin}}\left(\mathrm{4}\boldsymbol{\mathrm{x}}\right)<\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \left(\mathrm{4}\boldsymbol{\mathrm{x}}\right) \\ $$$$\boldsymbol{\mathrm{solve}}. \\ $$

Commented by bshahid010@gmail.com last updated on 03/Apr/19

1−2sin(4x)<cos^2 (4x) 1−2sin(4x)−cos^2 (4x)<0 1−2sin(4x)−1+sin^2 (4x)<0 sin^2 4x−2sin4x<0 sin4x(sin4x−2)<0 sin4x∈(0,2) so sin4x∈(0,1) x∈(0,sin1)

$$\mathrm{1}−\mathrm{2sin}\left(\mathrm{4x}\right)<\mathrm{cos}^{\mathrm{2}} \left(\mathrm{4x}\right) \\ $$$$\mathrm{1}−\mathrm{2sin}\left(\mathrm{4x}\right)−\mathrm{cos}^{\mathrm{2}} \left(\mathrm{4x}\right)<\mathrm{0} \\ $$$$\mathrm{1}−\mathrm{2sin}\left(\mathrm{4x}\right)−\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \left(\mathrm{4x}\right)<\mathrm{0} \\ $$$$\mathrm{sin}^{\mathrm{2}} \mathrm{4x}−\mathrm{2sin4x}<\mathrm{0} \\ $$$$\mathrm{sin4x}\left(\mathrm{sin4x}−\mathrm{2}\right)<\mathrm{0} \\ $$$$\mathrm{sin4x}\in\left(\mathrm{0},\mathrm{2}\right)\:\mathrm{so}\:\mathrm{sin4x}\in\left(\mathrm{0},\mathrm{1}\right) \\ $$$$\mathrm{x}\in\left(\mathrm{0},\mathrm{sin1}\right) \\ $$$$ \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 03/Apr/19

max value of sinθ=1 so i think sin4x∈(0,1) but sin4x∉(0,2) let other comment...

$${max}\:{value}\:{of}\:{sin}\theta=\mathrm{1}\: \\ $$$${so}\:{i}\:{think}\:\:\:\:{sin}\mathrm{4}{x}\in\left(\mathrm{0},\mathrm{1}\right)\:{but}\: \\ $$$${sin}\mathrm{4}{x}\notin\left(\mathrm{0},\mathrm{2}\right)\: \\ $$$${let}\:{other}\:{comment}… \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Apr/19

sin4x=a 1−2a<1−a^2 a^2 −2a<0 a(a−2)<0 a≠2 a_(max) =1 1>a>0 is the reauired solution 1>sin4x>0 sin(π/2)>sin4x>sin0 (π/8)>x>0

$${sin}\mathrm{4}{x}={a} \\ $$$$\mathrm{1}−\mathrm{2}{a}<\mathrm{1}−{a}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} −\mathrm{2}{a}<\mathrm{0} \\ $$$${a}\left({a}−\mathrm{2}\right)<\mathrm{0} \\ $$$${a}\neq\mathrm{2}\:\:\:\:\:\:\:{a}_{{max}} =\mathrm{1} \\ $$$$\mathrm{1}>{a}>\mathrm{0}\:{is}\:{the}\:{reauired}\:{solution}\: \\ $$$$\mathrm{1}>{sin}\mathrm{4}{x}>\mathrm{0} \\ $$$${sin}\frac{\pi}{\mathrm{2}}>{sin}\mathrm{4}{x}>{sin}\mathrm{0} \\ $$$$\frac{\pi}{\mathrm{8}}>{x}>\mathrm{0} \\ $$$$ \\ $$

Commented by ANTARES VY last updated on 03/Apr/19

$$??? \\ $$

Answered by einsteindrmaths@hotmail.fr last updated on 03/Apr/19

cos^2 (4x)=1−sin^2 (4x)====>1−sin(4x)<1−sin^2 (4x) ==>sin^2 (4x)−sin(4x)<0==>sin(4x)(1−sin(4x))<0==> sin(4x)>0&1−sin(4x)≠0===>x≠{(π/8)+((kπ)/2)/k∈IZ}∪]((kπ)/(2 ));(((2k+1)π)/4)[

$${c}\mathrm{os}\:^{\mathrm{2}} \left(\mathrm{4}{x}\right)=\mathrm{1}−{sin}^{\mathrm{2}} \left(\mathrm{4}{x}\right)====>\mathrm{1}−\mathrm{sin}\left(\mathrm{4}{x}\right)<\mathrm{1}−{sin}^{\mathrm{2}} \left(\mathrm{4}{x}\right) \\ $$$$==>{sin}^{\mathrm{2}} \left(\mathrm{4}{x}\right)−{sin}\left(\mathrm{4}{x}\right)<\mathrm{0}==>{sin}\left(\mathrm{4}{x}\right)\left(\mathrm{1}−{sin}\left(\mathrm{4}{x}\right)\right)<\mathrm{0}==> \\ $$$$\left.{sin}\left(\mathrm{4}{x}\right)>\mathrm{0\&1}−{sin}\left(\mathrm{4}{x}\right)\neq\mathrm{0}===>{x}\neq\left\{\frac{\pi}{\mathrm{8}}+\frac{{k}\pi}{\mathrm{2}}/{k}\in{IZ}\right\}\cup\right]\frac{{k}\pi}{\mathrm{2}\:};\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{4}}\left[\right. \\ $$

Answered by mr W last updated on 03/Apr/19

1−2 sin (4x)<1−sin^2 (4x) sin (4x){sin (4x)−2}<0 since sin (4x)−2<0 ⇒sin (4x)>0 or ⇒0<sin (4x)≤1 ⇒4x∈(2nπ,2nπ+π) ⇒x∈(((nπ)/2),((nπ)/2)+(π/4)) with n∈W

$$\mathrm{1}−\mathrm{2}\:\mathrm{sin}\:\left(\mathrm{4}{x}\right)<\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\left(\mathrm{4}{x}\right) \\ $$$$\mathrm{sin}\:\left(\mathrm{4}{x}\right)\left\{\mathrm{sin}\:\left(\mathrm{4}{x}\right)−\mathrm{2}\right\}<\mathrm{0} \\ $$$${since}\:\mathrm{sin}\:\left(\mathrm{4}{x}\right)−\mathrm{2}<\mathrm{0} \\ $$$$\Rightarrow\mathrm{sin}\:\left(\mathrm{4}{x}\right)>\mathrm{0} \\ $$$${or} \\ $$$$\Rightarrow\mathrm{0}<\mathrm{sin}\:\left(\mathrm{4}{x}\right)\leqslant\mathrm{1} \\ $$$$\Rightarrow\mathrm{4}{x}\in\left(\mathrm{2}{n}\pi,\mathrm{2}{n}\pi+\pi\right) \\ $$$$\Rightarrow{x}\in\left(\frac{{n}\pi}{\mathrm{2}},\frac{{n}\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\:{with}\:{n}\in\mathbb{W} \\ $$

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