document.write( "Question 471776: solve cosx(sinx-1/2)=0 [0-2/pi] \n" ); document.write( "
Algebra.Com's Answer #323533 by oadeshina(1)\"\" \"About 
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cosx*(sinx-1\2)=0
\n" ); document.write( "either cosx=0 or sinx-1\2=0
\n" ); document.write( "if sinx-1\2=0, then sinx=1\2,sinx=0.5, x=sin^1*0.5
\n" ); document.write( "then x=30. thanks
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