Question #0e470

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Answer 1 · 2018-01-05T08:52:00

$(2cosx+sin(3x))/((3-4sin^2x)cosx)$

$(2sinx)/sin(3x)+tanx/tan(3x)=(2sinx)/sin(3x)+(sinx/cosx)/(sin(3x)/cos(3x))=$

$=((2sinx)/cos(3x)+(sinx/cosx)*(sin(3x)/(cos(3x))) ) /(sin(3x)/cos(3x))=$

$=(((2sinxcosx+sinx*(sin(3x)))/(cancelcos(3x)*cosx)) ) /(sin(3x)/cancelcos(3x))=$

$=(( 2sinxcosx+sinx*(sin(3x)) )/cosx)/(sin(3x))=$

$= (2sinxcosx+sinx*sin(3x))/(sin(3x)cosx)=$

$sin(3x)=3sinx-4sin^3x$

$= (2sinxcosx+sinx*sin(3x))/((3sinx-4sin^3x)cosx)=$

$= (cancel(sinx)(2cosx+sin(3x)))/(cancel(sinx)(3-4sin^2x)cosx)=$

$= (2cosx+sin(3x))/((3-4sin^2x)cosx)$