document.write( "Question 445380: How do you solve: where Θ, 0° ≤ Θ < 360°
\n" ); document.write( " cos2Θ = cosΘ \n" ); document.write( "

Algebra.Com's Answer #306824 by htmentor(1343) $\"\"$ $\"About$

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$\"cos%282THETA%29+=+cos%28THETA%29\"$
\n" ); document.write( "Using the identity $\"cos%282THETA%29+=+2%28cos%28THETA%29%29%5E2+-+1\"$ we have
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\n" ); document.write( "This is a quadratic in $\"cos%28THETA%29\"$ . Let x = $\"cos%28THETA%29\"$ . Then we can write:
\n" ); document.write( " $\"2x%5E2+-+x+-+1+=+0\"$
\n" ); document.write( "Solve for x using the quadratic formula:
\n" ); document.write( " $\"x+=+%281+%2B-+sqrt%281+%2B+8%29%29%2F4\"$
\n" ); document.write( "This gives x = 1, -1/2
\n" ); document.write( "Since $\"x+=+cos%28THETA%29\"$ we need to find values for $\"THETA\"$ which give $\"cos%28THETA%29+=+1\"$ , $\"-1%2F2\"$
\n" ); document.write( "The inv. cos of 1 = 0 deg.
\n" ); document.write( "The inv. cos of -1/2 = 120 deg.
\n" ); document.write( "Since the $\"THETA\"$ ranges from 0 to 360 deg. and since $\"2THETA\"$ = 240 deg., we need to include 240 deg.
\n" ); document.write( "So the answers are $\"THETA+=+0\"$ deg, $\"THETA+=+120\"$ deg, $\"THETA+=+240\"$ deg \n" ); document.write( "

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