document.write( "Question 971540: Solve on the interval [0, 2pi]
\n" ); document.write( "Sin*2x - sin x + 1= cos*2x \n" ); document.write( "

Algebra.Com's Answer #594185 by lwsshak3(11628) $\"\"$ $\"About$

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Solve on the interval [0, 2pi]
\n" ); document.write( "Sin*2x - sin x + 1= cos*2x
\n" ); document.write( "sin^2(x)-sinx+1=1-sin^2(x)
\n" ); document.write( "2sin^2(x)-sinx=0
\n" ); document.write( "sinx(2sinx-1)=0
\n" ); document.write( "sinx=0
\n" ); document.write( "x=0, π, 2π
\n" ); document.write( "or
\n" ); document.write( "2sinx-1=0
\n" ); document.write( "sinx=1/2
\n" ); document.write( "x=π/6, 11π/6
\n" ); document.write( " \n" ); document.write( "

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