document.write( "Question 743225: 2sin^2(u)=−1−3sin(u)\r
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document.write( "Find the solutions of the equation that are in the interval [0, 2π).\r
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document.write( "Please help! \n" );
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Algebra.Com's Answer #452820 by lwsshak3(11628)![]() ![]() ![]() You can put this solution on YOUR website! 2sin^2(u)=−1−3sin(u) \n" ); document.write( "Find the solutions of the equation that are in the interval [0, 2π). \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "(2sin(u)+1)(sin(u)+1)=0 \n" ); document.write( ".. \n" ); document.write( "2sin(u)+1=0 \n" ); document.write( "sin(u)=-1/2 \n" ); document.write( "u=7π/6, 11π/6 \n" ); document.write( ".. \n" ); document.write( "sin(u)+1=0 \n" ); document.write( "sin(u)=-1 \n" ); document.write( "u=3π/2 \n" ); document.write( " |