document.write( "Question 359681: Solve the equation algebraically for \"x\" where 0≤x<2π.
\n" ); document.write( "sinx + cosx = 1\r
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\n" ); document.write( "\n" ); document.write( "Thank you \n" ); document.write( "

Algebra.Com's Answer #256675 by Fombitz(32388) $\"\"$ $\"About$

You can put this solution on YOUR website!
$\"sin%28x%29%2Bcos%28x%29=1\"$
\n" ); document.write( " $\"sin%28x%29=1-cos%28x%29\"$
\n" ); document.write( " $\"%28sin%28x%29%29%5E2=%281-cos%28x%29%29%5E2\"$
\n" ); document.write( " $\"1-%28cos%28x%29%29%5E2=1-2cos%28x%29%2Bcos%28x%29%5E2\"$
\n" ); document.write( " $\"2%28cos%28x%29%29%5E2-2cos%28x%29=0\"$
\n" ); document.write( " $\"cos%28x%29%2A%28cos%28x%29-1%29=0\"$
\n" ); document.write( "Two solutions:
\n" ); document.write( " $\"cos%28x%29=0\"$
\n" ); document.write( " $\"x=pi%2F2\"$ and $\"x=%283pi%29%2F2\"$
\n" ); document.write( "However at $\"x=%283pi%29%2F2\"$ , $\"sin%28x%29=-1\"$ and $\"cos%28x%29=0\"$
\n" ); document.write( "so that $\"sin%28x%29%2Bcos%28x%29%3C%3E1\"$
\n" ); document.write( "Only $\"highlight%28x=pi%2F2%29\"$ is a valid solution.
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\n" ); document.write( ".
\n" ); document.write( " $\"cos%28x%29-1=0\"$
\n" ); document.write( " $\"cos%28x%29=1\"$
\n" ); document.write( " $\"highlight%28x=0%29\"$
\n" ); document.write( " \n" ); document.write( "

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