document.write( "Question 1133311: If $\"%28sinA%2BcosA%29%2FsinAcosA=1%2F3\"$
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Algebra.Com's Answer #750610 by ikleyn(52909) $\"\"$ $\"About$

You can put this solution on YOUR website!
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\n" ); document.write( "\n" ); document.write( "I will solve it in two steps.\r
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document.write( "For brevity, let's denote x = sin(A)*cos(A).\r\n" );
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document.write( " =   ====>  square both sides  ====>  \r\n" );
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document.write( " =  =   ====>\r\n" );
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document.write( "9*(1 + 2x) = \r\n" );
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document.write( " = 0\r\n" );
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document.write( " =  = .\r\n" );
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document.write( "Since the modulus of x,  |x|, must be less than 1, only the root   x =   is the solution.\r\n" );
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document.write( "It implies  2x = 2*sin(A)*cos(A) = sin(2A) =  = .      (1)\r\n" );
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document.write( " = .  ====>  replace    by 1  ====>\r\n" );
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document.write( " =  =  = .\r\n" );
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document.write( "Substitute here   from (1)  and continue\r\n" );
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document.write( " =  =  =  =  = \r\n" );
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document.write( "=  = -0.228 (approximately, with 3 correct decimal places).\r\n" );
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document.write( "ANSWER.  If   = ,  then   =  = -0.228 (approximately, with 3 correct decimal places).\r\n" );
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