document.write( "Question 1139747: Please help me solve this problem:
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document.write( "cot-1[tan(2x)] + cot-1[tan(3x)] = x \n" );
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Algebra.Com's Answer #760292 by Edwin McCravy(20056)![]() ![]() You can put this solution on YOUR website! \r\n" ); document.write( "cot-1[tan(2x)] + cot-1[tan(3x)] = x\r\n" ); document.write( "\r\n" ); document.write( "Since tangent and cotangent have period 180°, \r\n" ); document.write( "\r\n" ); document.write( "cot-1[tan(2x±180°k)] + cot-1[tan(3x±180°)] = x\r\n" ); document.write( "\r\n" ); document.write( "replace tan(2x±180°k) by cot[90°-(2x±180°k)] and tan(3x±180°k) by \r\n" ); document.write( "cot[90°-(3x+180°k)]\r\n" ); document.write( "\r\n" ); document.write( "cot-1{cot[90°-(2x±180°)]} + cot-1{cot[90°-(3x±180°)]} = x\r\n" ); document.write( "\r\n" ); document.write( "[90°-(2x±180°k)] + [90°-(3x±180°k)] = x\r\n" ); document.write( "\r\n" ); document.write( "90° - 2x ∓ 180°k + 90° - 3x ∓ 180°k = x\r\n" ); document.write( "\r\n" ); document.write( " 180° ∓ 360°k - 5x = x\r\n" ); document.write( "\r\n" ); document.write( " 180° ∓ 360°k = 6x\r\n" ); document.write( " \r\n" ); document.write( " 180°(1 ∓ 2k) = 6x\r\n" ); document.write( "\r\n" ); document.write( " 30°(1 ∓ 2k) = x\r\n" ); document.write( "\r\n" ); document.write( "Since k can take on negative integer values, we\r\n" ); document.write( "can just use a + sign.\r\n" ); document.write( "\r\n" ); document.write( " 30°(2k + 1) = x\r\n" ); document.write( "\r\n" ); document.write( "So x can be any odd multiple of 30°.\r\n" ); document.write( "\r\n" ); document.write( "However we must discard this because the original equation contains\r\n" ); document.write( "\r\n" ); document.write( "tan(3x), which would be undefined, for 3x would be odd multiples of 90°\r\n" ); document.write( "and tangents of odd multiples of 90° is undefined.\r\n" ); document.write( "\r\n" ); document.write( "So after all that, we find that there is no solution.\r\n" ); document.write( "\r\n" ); document.write( "Edwin\n" ); document.write( " |