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Question Number 131622 by rydasss last updated on 06/Feb/21

for -π < x < - (π/2) find the solution of sin x > ((sin x + 2)/(2 sin x + 1))

$${for}\:-\pi\:<\:{x}\:<\:-\:\frac{\pi}{\mathrm{2}}\:{find}\:{the}\:{solution}\:{of} \\ $$ $${sin}\:{x}\:>\:\:\frac{{sin}\:{x}\:+\:\mathrm{2}}{\mathrm{2}\:{sin}\:{x}\:+\:\mathrm{1}} \\ $$

Answered by mr W last updated on 07/Feb/21

−π<x<−(π/2) ⇒ −1<sin x<0 case 1: 2sin x+1>0, i.e. sin x>−(1/2) sin x(2sin x+1)>sin x+2 sin^2 x>1 ⇒no solution! case 2: 2sin x+1<0, i.e. sin x<−(1/2) sin x(2sin x+1)<sin x+2 sin^2 x<1 ⇒sin x>−1 ⇒−((5π)/6)<x<−(π/2) (solution)

$$−\pi<{x}<−\frac{\pi}{\mathrm{2}}\:\Rightarrow\:−\mathrm{1}<\mathrm{sin}\:{x}<\mathrm{0} \\ $$ $${case}\:\mathrm{1}:\:\mathrm{2sin}\:{x}+\mathrm{1}>\mathrm{0},\:{i}.{e}.\:\mathrm{sin}\:{x}>−\frac{\mathrm{1}}{\mathrm{2}} \\ $$ $$\mathrm{sin}\:{x}\left(\mathrm{2sin}\:{x}+\mathrm{1}\right)>\mathrm{sin}\:{x}+\mathrm{2} \\ $$ $$\mathrm{sin}^{\mathrm{2}} \:{x}>\mathrm{1} \\ $$ $$\Rightarrow{no}\:{solution}! \\ $$ $${case}\:\mathrm{2}:\:\mathrm{2sin}\:{x}+\mathrm{1}<\mathrm{0},\:{i}.{e}.\:\mathrm{sin}\:{x}<−\frac{\mathrm{1}}{\mathrm{2}} \\ $$ $$\mathrm{sin}\:{x}\left(\mathrm{2sin}\:{x}+\mathrm{1}\right)<\mathrm{sin}\:{x}+\mathrm{2} \\ $$ $$\mathrm{sin}^{\mathrm{2}} \:{x}<\mathrm{1} \\ $$ $$\Rightarrow\mathrm{sin}\:{x}>−\mathrm{1} \\ $$ $$\Rightarrow−\frac{\mathrm{5}\pi}{\mathrm{6}}<{x}<−\frac{\pi}{\mathrm{2}}\:\left({solution}\right) \\ $$

Commented byrydasss last updated on 07/Feb/21

Thank you so much.

$${Thank}\:{you}\:{so}\:{much}. \\ $$

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