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))](https://www.tinkutara.com/question/Q131622.png)
$${for}\:-\pi\:<\:{x}\:<\:-\:\frac{\pi}{\mathrm{2}}\:{find}\:{the}\:{solution}\:{of} \\ $$$${sin}\:{x}\:>\:\:\frac{{sin}\:{x}\:+\:\mathrm{2}}{\mathrm{2}\:{sin}\:{x}\:+\:\mathrm{1}} \\ $$
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)](https://www.tinkutara.com/question/Q131623.png)
$$−\pi<{x}<−\frac{\pi}{\mathrm{2}}\:\Rightarrow\:−\mathrm{1}<\mathrm{sin}\:{x}<\mathrm{0} \\ $$$${case}\:\mathrm{1}:\:\mathrm{2sin}\:{x}+\mathrm{1}>\mathrm{0},\:{i}.{e}.\:\mathrm{sin}\:{x}>−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{sin}\:{x}\left(\mathrm{2sin}\:{x}+\mathrm{1}\right)>\mathrm{sin}\:{x}+\mathrm{2} \\ $$$$\mathrm{sin}^{\mathrm{2}} \:{x}>\mathrm{1} \\ $$$$\Rightarrow{no}\:{solution}! \\ $$$${case}\:\mathrm{2}:\:\mathrm{2sin}\:{x}+\mathrm{1}<\mathrm{0},\:{i}.{e}.\:\mathrm{sin}\:{x}<−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{sin}\:{x}\left(\mathrm{2sin}\:{x}+\mathrm{1}\right)<\mathrm{sin}\:{x}+\mathrm{2} \\ $$$$\mathrm{sin}^{\mathrm{2}} \:{x}<\mathrm{1} \\ $$$$\Rightarrow\mathrm{sin}\:{x}>−\mathrm{1} \\ $$$$\Rightarrow−\frac{\mathrm{5}\pi}{\mathrm{6}}<{x}<−\frac{\pi}{\mathrm{2}}\:\left({solution}\right) \\ $$
Commented by rydasss last updated on 07/Feb/21
![Thank you so much.](https://www.tinkutara.com/question/Q131632.png)
$${Thank}\:{you}\:{so}\:{much}. \\ $$