document.write( "Question 518554: Solve: cos-1(x) + cos-1(2x) = 60°
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Algebra.Com's Answer #345231 by Edwin McCravy(20056)\"\" \"About 
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document.write( "cos-1(x) + cos-1(2x) = 60°\r\n" );
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document.write( "Let \"alpha\" = cos-1(x)\r\n" );
document.write( "Let \"beta\" = cos-1(2x)\r\n" );
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document.write( "Then cos(\"alpha\") = x\r\n" );
document.write( "and cos(\"beta\") = 2x\r\n" );
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document.write( "So the equation\r\n" );
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document.write( "cos-1(x) + cos-1(2x) = 60°\r\n" );
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document.write( "becomes \r\n" );
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document.write( "\"alpha\" + \"beta\" = 60°\r\n" );
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document.write( "Take the cosine of both sides:\r\n" );
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document.write( "cos(\"alpha\"+\"beta\") = cos(60°)\r\n" );
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document.write( "Using an identity for cos(\"alpha\"+\"beta\"),\r\n" );
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document.write( "cos(\"alpha\")cos(\"beta\") - sin(\"alpha\")sin(\"beta\") = \"1%2F2\"\r\n" );
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document.write( "Then since cos(\"alpha\") = x and cos(\"beta\") = 2x\r\n" );
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document.write( "x(2x) - sin(\"alpha\")sin(\"beta\") = \"1%2F2\"\r\n" );
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document.write( "2x² - sin(\"alpha\")sin(\"beta\") = \"1%2F2\"\r\n" );
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document.write( "Now we have to find the sines.  We use the identity:\r\n" );
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document.write( "sin²\"theta\" + cos²\"theta\" = 1\r\n" );
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document.write( "sin²\"theta\" = 1 - cos²\"theta\" \r\n" );
document.write( "          ___________ \r\n" );
document.write( "sin\"theta\" = ±Ö1 - cos²\"theta\"\r\n" );
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document.write( "The sign of the sine depends on what quadrants \"alpha\" and \"beta\" are in.  We consider all cases\r\n" );
document.write( "           ___________     ______\r\n" );
document.write( "sin(a) = ±Ö1 - cos²(a) = ±Ö1 - x²\r\n" );
document.write( "           ___________     __________    _________ \r\n" );
document.write( "sin(b) = ±Ö1 - cos²(b} = ±Ö1 - (2x)² = ±Ö1 - 4x²\r\n" );
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document.write( "Substituting in\r\n" );
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document.write( "2x² - sin(\"alpha\")sin(\"beta\") = \"1%2F2\"\r\n" );
document.write( "       ______  _______\r\n" );
document.write( "2x² ± Ö1 - x²·Ö1 - 4x² = \"1%2F2\"\r\n" );
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document.write( "       _________________\r\n" );
document.write( "2x² ± Ö(1 - x²)(1 - 4x²) = \"1%2F2\"\r\n" );
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document.write( "Multiply both sides by 2 to clear of the fraction:\r\n" );
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document.write( "        _________________\r\n" );
document.write( "4x² ± 2Ö(1 - x²)(1 - 4x²) = 1\r\n" );
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document.write( "Isolate the radical term:\r\n" );
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document.write( "         ___________________\r\n" );
document.write( "      ±2Ö(1 - x²)(1 - 4x²) = 1 - 4x²\r\n" );
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document.write( "Square both sides:\r\n" );
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document.write( "        4(1 - x²)(1 - 4x²) = (1 - 4x²)²\r\n" );
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document.write( "          4(1 - 5x² + 4x4) = 1 - 8x² + 16x4\r\n" );
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document.write( "           4 - 20x² + 16x4 = 1 - 8x² + 16x4\r\n" );
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document.write( "                     -12x² = -3        \r\n" );
document.write( "                          \r\n" );
document.write( "                        x² = \"%28-3%29%2F%28-12%29\"\r\n" );
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document.write( "                        x² = \"1%2F4\"\r\n" );
document.write( "                          \r\n" );
document.write( "                         x = ±\"1%2F2\"\r\n" );
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document.write( "We must check for extraneous solutions:\r\n" );
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document.write( "Checking \"1%2F2\":\r\n" );
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document.write( "cos-1(x) + cos-1(2x) = 60°\r\n" );
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document.write( "cos-1(\"1%2F2\") + cos-1(2·\"1%2F2\") = 60°\r\n" );
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document.write( "60° + cos-1(1) = 60°\r\n" );
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document.write( "60° + 0° = 60°\r\n" );
document.write( "   \r\n" );
document.write( "60° = 60°\r\n" );
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document.write( "That checks, so \"1%2F2\" is a solution.\r\n" );
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document.write( "Checking \"-1%2F2\":\r\n" );
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document.write( "cos-1(x) + cos-1(2x) = 60°\r\n" );
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document.write( "cos-1(\"-1%2F2\") + cos-1(2·\"-1%2F2\") = 60°\r\n" );
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document.write( "120° + cos-1(-1) = 60°\r\n" );
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document.write( "120° + 180° = 60°\r\n" );
document.write( "   \r\n" );
document.write( "300° = 60°\r\n" );
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document.write( "That's false, so the only solution is \r\n" );
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document.write( "x = \"1%2F2\"\r\n" );
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document.write( "Edwin
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