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Question Number 193366 by aipman last updated on 11/Jun/23

2sin^2 2x>3cos x+3

$$\mathrm{2sin}\:^{\mathrm{2}} \mathrm{2}{x}>\mathrm{3cos}\:{x}+\mathrm{3} \\ $$

Answered by Frix last updated on 11/Jun/23

2sin^2 2x −3(1+cos x)>0 c=cos x −8c^4 +8c^2 −3c−3>0 c^4 −c^2 +((3c)/8)+(3/8)<0 (c+1)(c+(1/2))(c^2 −((3c)/2)+(3/4))<0 c^2 −((3c)/2)+(3/4)≥(3/(16)) ⇒ c+1<0∧c+(1/2)>0 ∨ c+1>0∧c+(1/2)<0 c<−1 not possible ⇒ −1<c<−(1/2) −1<cos x <−(1/2) The rest is easy

$$\mathrm{2sin}^{\mathrm{2}} \:\mathrm{2}{x}\:−\mathrm{3}\left(\mathrm{1}+\mathrm{cos}\:{x}\right)>\mathrm{0} \\ $$$${c}=\mathrm{cos}\:{x} \\ $$$$−\mathrm{8}{c}^{\mathrm{4}} +\mathrm{8}{c}^{\mathrm{2}} −\mathrm{3}{c}−\mathrm{3}>\mathrm{0} \\ $$$${c}^{\mathrm{4}} −{c}^{\mathrm{2}} +\frac{\mathrm{3}{c}}{\mathrm{8}}+\frac{\mathrm{3}}{\mathrm{8}}<\mathrm{0} \\ $$$$\left({c}+\mathrm{1}\right)\left({c}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left({c}^{\mathrm{2}} −\frac{\mathrm{3}{c}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{4}}\right)<\mathrm{0} \\ $$$${c}^{\mathrm{2}} −\frac{\mathrm{3}{c}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{4}}\geqslant\frac{\mathrm{3}}{\mathrm{16}} \\ $$$$\Rightarrow \\ $$$${c}+\mathrm{1}<\mathrm{0}\wedge{c}+\frac{\mathrm{1}}{\mathrm{2}}>\mathrm{0}\:\vee\:{c}+\mathrm{1}>\mathrm{0}\wedge{c}+\frac{\mathrm{1}}{\mathrm{2}}<\mathrm{0} \\ $$$${c}<−\mathrm{1}\:\mathrm{not}\:\mathrm{possible} \\ $$$$\Rightarrow \\ $$$$−\mathrm{1}<{c}<−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$−\mathrm{1}<\mathrm{cos}\:{x}\:<−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{The}\:\mathrm{rest}\:\mathrm{is}\:\mathrm{easy} \\ $$

Commented by aipman last updated on 13/Jun/23

≥(3/(16)) where is it comes from?

$$ \\ $$$$\geqslant\frac{\mathrm{3}}{\mathrm{16}}\:\mathrm{where}\:\mathrm{is}\:\mathrm{it}\:\mathrm{comes}\:\mathrm{from}? \\ $$

Commented by Frix last updated on 16/Jun/23

The minimum of f(c)=c^2 −((3c)/2)+(3/4) is (3/(16))

$$\mathrm{The}\:\mathrm{minimum}\:\mathrm{of}\:{f}\left({c}\right)={c}^{\mathrm{2}} −\frac{\mathrm{3}{c}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{4}}\:\mathrm{is}\:\frac{\mathrm{3}}{\mathrm{16}} \\ $$

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