document.write( "Question 794887: Please help me solve this equation: $\"+cosx=sin2x+\"$ . \n" ); document.write( "

Algebra.Com's Answer #480771 by lwsshak3(11628) $\"\"$ $\"About$

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[0,2π]
\n" ); document.write( "solve this equation: $\"+cosx=sin2x+\"$ .
\n" ); document.write( "cosx=sin2x
\n" ); document.write( "cosx-sin2x=0
\n" ); document.write( "cosx-2sinxcosx=0
\n" ); document.write( "cosx(1-2sinx)=0
\n" ); document.write( "cosx=0
\n" ); document.write( "x=π/2, 3π/2
\n" ); document.write( "..
\n" ); document.write( "1-2sinx=0
\n" ); document.write( "2sinx=1
\n" ); document.write( "sinx=1/2
\n" ); document.write( "x=π/6, 5π/6
\n" ); document.write( "..
\n" ); document.write( "x=π/6, π/2, 5π/6, 3π/2 \n" ); document.write( "

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