document.write( "Question 937255: Solve the following for 0 < x < 2pi\r
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Algebra.Com's Answer #854647 by ikleyn(53937) $\"\"$ $\"About$

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\n" ); document.write( "\n" ); document.write( "        In his post, @lwsshak3 came to conclusion \"no solution: x not in given interval.\"\r
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\n" ); document.write( "\n" ); document.write( "        This answer is INCORRECT. Given equation has 3 (three) solutions in the given interval.\r
\n" ); document.write( "\n" ); document.write( "        They are $\"pi%2F2\"$ , $\"pi\"$ and $\"3pi%2F2\"$ . \r
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\n" ); document.write( "\n" ); document.write( "        The reasoning in the post by @l2sshak3 is conceptually wrong, \r
\n" ); document.write( "\n" ); document.write( "        it is why he missed 3 solutions.\r
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\n" ); document.write( "\n" ); document.write( "        It is wrong way to teach students, so I came to bring a correct solution.\r
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document.write( "Your starting equation is\r\n" );
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document.write( "     = 1\r\n" );
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document.write( "Transform it step by step\r\n" );
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document.write( "     = 1,\r\n" );
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document.write( "    1 + 2*sin(x)*cos(x) = 1,\r\n" );
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document.write( "    sin(2x) = 0.\r\n" );
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document.write( "Hence, 2x should be a multiple of .\r\n" );
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document.write( "Since x should be in interval (,),  2x should be in interval (,)\r\n" );
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document.write( "    0 < 2x < .\r\n" );
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document.write( "There are 3 possible values for 2x: ,  and .\r\n" );
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document.write( "It gives 3 possible values for x: ,  and .\r\n" );
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document.write( "It is easy to check that all these 3 values satisfy given equation.\r\n" );
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document.write( "ANSWER.  Given equation has 3 solutions in the given interval: ,  and .\r\n" );
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\n" ); document.write( "\n" ); document.write( "Solved correctly to teach you in a right way.\r
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