document.write( "Question 1015049: Find all x in the interval (0, pi) that satisfy:\r
\n" ); document.write( "\n" ); document.write( "cot(x/2)>1+cot(x) \n" ); document.write( "

Algebra.Com's Answer #631383 by robertb(5830) $\"\"$ $\"About$

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We know that $\"tanx+=+%282tan%28x%2F2%29%29%2F%281-%28tan%28x%2F2%29%29%5E2%29\"$ , and so by inverting,
\n" ); document.write( " $\"cotx+=+%281-%28tan%28x%2F2%29%29%5E2%29%2F%282tan%28x%2F2%29%29\"$ .\r
\n" ); document.write( "\n" ); document.write( "The original inequality then becomes\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " $\"1%2Ftan%28x%2F2%29+%3E1+%2B+%281-%28tan%28x%2F2%29%29%5E2%29%2F%282tan%28x%2F2%29%29\"$ , or\r
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\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " $\"2%2F%282tan%28x%2F2%29%29+%3E1+%2B+%281-%28tan%28x%2F2%29%29%5E2%29%2F%282tan%28x%2F2%29%29\"$ .\r
\n" ); document.write( "\n" ); document.write( "==> $\"%281%2B%28tan%28x%2F2%29%29%5E2%29%2F%282tan%28x%2F2%29%29%3E1\"$ \r
\n" ); document.write( "\n" ); document.write( "The fact that x is in (0,pi) means that x/2 is in (0, pi/2), in which case tan(x/2) is always POSITIVE and so the last inequality becomes
\n" ); document.write( " $\"1%2B%28tan%28x%2F2%29%29%5E2+%3E+2tan%28x%2F2%29\"$ .\r
\n" ); document.write( "\n" ); document.write( "==> $\"1-2tan%28x%2F2%29%2B%28tan%28x%2F2%29%29%5E2+%3E0+\"$ , or $\"%281-tan%28x%2F2%29%29%5E2+%3E+0\"$ .\r
\n" ); document.write( "\n" ); document.write( "The last inequality is always true except when 1 - tan(x/2) = 0. This happens only when $\"x%2F2+=+pi%2F4\"$ , or $\"x+=+pi%2F2\"$ .\r
\n" ); document.write( "\n" ); document.write( "Therefore, the solution set of the inequality is (0,pi/2) U (pi/2,pi). \n" ); document.write( "

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