Question Number 131622 by rydasss last updated on 06/Feb/21 | ||
$${for}\:-\pi\:<\:{x}\:<\:-\:\frac{\pi}{\mathrm{2}}\:{find}\:{the}\:{solution}\:{of} \\ $$ $${sin}\:{x}\:>\:\:\frac{{sin}\:{x}\:+\:\mathrm{2}}{\mathrm{2}\:{sin}\:{x}\:+\:\mathrm{1}} \\ $$ | ||
Answered by mr W last updated on 07/Feb/21 | ||
$$−\pi<{x}<−\frac{\pi}{\mathrm{2}}\:\Rightarrow\:−\mathrm{1}<\mathrm{sin}\:{x}<\mathrm{0} \\ $$ $${case}\:\mathrm{1}:\:\mathrm{2sin}\:{x}+\mathrm{1}>\mathrm{0},\:{i}.{e}.\:\mathrm{sin}\:{x}>−\frac{\mathrm{1}}{\mathrm{2}} \\ $$ $$\mathrm{sin}\:{x}\left(\mathrm{2sin}\:{x}+\mathrm{1}\right)>\mathrm{sin}\:{x}+\mathrm{2} \\ $$ $$\mathrm{sin}^{\mathrm{2}} \:{x}>\mathrm{1} \\ $$ $$\Rightarrow{no}\:{solution}! \\ $$ $${case}\:\mathrm{2}:\:\mathrm{2sin}\:{x}+\mathrm{1}<\mathrm{0},\:{i}.{e}.\:\mathrm{sin}\:{x}<−\frac{\mathrm{1}}{\mathrm{2}} \\ $$ $$\mathrm{sin}\:{x}\left(\mathrm{2sin}\:{x}+\mathrm{1}\right)<\mathrm{sin}\:{x}+\mathrm{2} \\ $$ $$\mathrm{sin}^{\mathrm{2}} \:{x}<\mathrm{1} \\ $$ $$\Rightarrow\mathrm{sin}\:{x}>−\mathrm{1} \\ $$ $$\Rightarrow−\frac{\mathrm{5}\pi}{\mathrm{6}}<{x}<−\frac{\pi}{\mathrm{2}}\:\left({solution}\right) \\ $$ | ||
Commented byrydasss last updated on 07/Feb/21 | ||
$${Thank}\:{you}\:{so}\:{much}. \\ $$ | ||