document.write( "Question 445384: How do you solve cos2Θ = cosΘ where Θ, 0° ≤ Θ < 360°
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Algebra.Com's Answer #306816 by poliphob3.14(115) $\"\"$ $\"About$

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For convenience substitute $\"theta=x\"$ and modify the trigonometric equation:\r
\n" ); document.write( "\n" ); document.write( " $\"cos2x=cosx\"$ , since $\"cos2x=%28cosx%29%5E2-%28sinx%29%5E2\"$ and \r
\n" ); document.write( "\n" ); document.write( " $\"%28sinx%29%5E2=1-%28cosx%29%5E2\"$ , substituting we get: $\"%28cosx%29%5E2-%28sinx%29%5E2=cosx\"$ =>\r
\n" ); document.write( "\n" ); document.write( " $\"%28cosx%29%5E2-1%2B%28cosx%29%5E2=cosx\"$ => $\"2%28cosx%29%5E2-cosx-1=0\"$ . In the last equation\r
\n" ); document.write( "\n" ); document.write( "substitute $\"cosx=y\"$ and get the quadratic equation: $\"2y%5E2-y-1=0\"$ ,\r
\n" ); document.write( "\n" ); document.write( "Solving this equation $\"2y%5E2-y-1=%282y%2B1%29%28y-1%29=0\"$ we find y=1 and y=-1/2.\r
\n" ); document.write( "\n" ); document.write( "Trigonometric equation $\"cos2x=cosx\"$ is equivalent with two new equations:\r
\n" ); document.write( "\n" ); document.write( " $\"cosx=-1\"$ and $\"cosx=-1%2F2\"$ . Solving these equations we have;\r
\n" ); document.write( "\n" ); document.write( " $\"cosx=-1\"$ => $\"x=180\"$ degree and $\"cosx=-1%2F2\"$ => $\"x=120\"$ , and \r
\n" ); document.write( "\n" ); document.write( " $\"x=240\"$ degree. \n" ); document.write( "

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