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

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

Answered by Frix last updated on 11/Jun/23

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

Commented by aipman last updated on 13/Jun/23

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

Commented by Frix last updated on 16/Jun/23

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

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