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