Question Number 147936 by mathdanisur last updated on 24/Jul/21
$${Solve}\:{the}\:{inequality} \\ $$$$\left(\pi\:-\:\mathrm{3}\right)^{\mathrm{2}\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{x}}} \:\leqslant\:\left(\pi\:-\:\mathrm{3}\right)^{\mathrm{2}} \\ $$
Answered by iloveisrael last updated on 24/Jul/21
$$\Rightarrow\pi−\mathrm{3}\:>\mathrm{1}\:\mathrm{then}\:\mathrm{2sin}\:\mathrm{2x}\:\leqslant\mathrm{2} \\ $$$$\Rightarrow\mathrm{sin}\:\mathrm{2x}\:\leqslant\mathrm{1}\: \\ $$$$\Rightarrow\forall\mathrm{x}\in\mathrm{R}\: \\ $$$$ \\ $$
Commented by mathdanisur last updated on 24/Jul/21
$${Thank}\:{you}\:{Ser},\:{but}\:{answer}.? \\ $$