Algebra.Com's Answer #775067 by Edwin McCravy(20060)  You can put this solution on YOUR website! By substituting x=2cosy into cos3y= 4cos^3y - 3cosy , show that the equation \n" );
document.write( "(x)^3 - 3x - 1 = 0 has roots 2cos20, -2sin10 and -2cos40 \n" );
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document.write( "So we substitute \r\n" );
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document.write( "for the first two terms in\r\n" );
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document.write( "and get\r\n" );
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document.write( "We'll just get the answers for y in [0°,360°).\r\n" );
document.write( "That means we get the answers for 3y in [0°,1080°)\r\n" );
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document.write( "Since x = 2cos(y),\r\n" );
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document.write( "and since cos(20°)=cos(340°), cos(140°)=cos(220°), and cos(100°)=cos(260°),\r\n" );
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document.write( "we only have 3 solutions for x,\r\n" );
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document.write( "2cos(20°), 2cos(100°), 2cos(140°)\r\n" );
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document.write( "The first solution is already in the given form. The other two are not, so\r\n" );
document.write( "we'll need to work on them. For the second solution, we use cos(θ) = sin(90°-θ),\r\n" );
document.write( "and sin(-θ) = -sin(θ):\r\n" );
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document.write( "2cos(100°) = 2sin(90°-100°) = 2sin(-10°) = -2sin(10°) \r\n" );
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document.write( "For the third solution, we use cos(θ) = -cos(180°-θ)\r\n" );
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document.write( "2cos(140°) = -2cos(180°-140°) = -2cos(40°) \r\n" );
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document.write( "Edwin \n" );
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