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Question Number 16085 by Tinkutara last updated on 18/Jun/17
$$\mathrm{Let}\:{f}\left(\mathrm{sin}\:{x}\right)\:+\:\mathrm{2}{f}\left(\mathrm{cos}\:{x}\right)\:=\:\mathrm{3}\:\forall\:{x}\:\in\:\left(\mathrm{0},\:\frac{\pi}{\mathrm{2}}\right). \\ $$$$\mathrm{Then} \\ $$$$\left(\mathrm{1}\right)\:{f}\left(\mathrm{sin}\:{x}\right)\:=\:\mathrm{1},\:{x}\:\in\:\left(\mathrm{0},\:\frac{\pi}{\mathrm{2}}\right) \\ $$$$\left(\mathrm{2}\right)\:{f}\left(\mathrm{sin}\:{x}\right)\:=\:\mathrm{1},\:{x}\:\in\:\left(−\mathrm{1},\:\mathrm{0}\right) \\ $$$$\left(\mathrm{3}\right)\:{f}\left(\mathrm{cos}\:{x}\right)\:=\:\mathrm{1},\:{x}\:\in\:\left(\mathrm{0},\:\mathrm{1}\right) \\ $$$$\left(\mathrm{4}\right)\:{f}\left({x}\right)\:=\:\mathrm{1},\:{x}\:\in\:\left(\mathrm{0},\:\mathrm{1}\right) \\ $$
Commented by prakash jain last updated on 19/Jun/17
$${f}\left(\mathrm{sin}\:{x}\right)+\mathrm{2}{f}\left(\mathrm{cos}\:{x}\right)=\mathrm{3} \\ $$$${x}=\frac{\pi}{\mathrm{2}}−{x} \\ $$$${f}\left(\mathrm{cos}\:{x}\right)+\mathrm{2}{f}\left(\mathrm{sin}\:{x}\right)=\mathrm{3} \\ $$$${f}\left(\mathrm{cos}\:{x}\right)−{f}\left(\mathrm{sin}\:{x}\right)= \\ $$$$\Rightarrow{f}\left(\mathrm{cos}\:{x}\right)={f}\left(\mathrm{sin}\:{x}\right) \\ $$$$\Rightarrow\mathrm{3}{f}\left(\mathrm{cos}\:{x}\right)=\mathrm{3}\Rightarrow{f}\left(\mathrm{cos}\:{x}\right)=\mathrm{1}={f}\left(\mathrm{sin}\:{x}\right) \\ $$$${f}\left(\mathrm{sin}\:{x}\right)=\mathrm{1}={f}\left(\mathrm{cos}{x}\right)\:,{x}\in\left(\mathrm{0},\frac{\pi}{\mathrm{2}}\right) \\ $$$$\Rightarrow{f}\left({x}\right)=\mathrm{1},\:{x}\in\left(\mathrm{0},\mathrm{1}\right) \\ $$