document.write( "Question 630941: how do you solve sin(x/2)+cosx=0 with a restriction of [0,2pi)? \n" ); document.write( "

Algebra.Com's Answer #451009 by Barbazzo(2) $\"\"$ $\"About$

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$\"sin%28x%2F2%29%2Bcos+x+=+0\"$
\n" ); document.write( "Transform Cos into Sin using Double Angle Identity: $\"cos%282u%29+=+1-2sin%5E2%28u%29\"$
\n" ); document.write( "Let $\"u+=+x%2F2\"$ , so $\"cos%28x%29+=+1-2sin%5E2%28x%2F2%29\"$
\n" ); document.write( " $\"sin%28x%2F2%29%2B1-2sin%5E2%28x%2F2%29+=+0\"$
\n" ); document.write( " $\"-2sin%5E2%28x%2F2%29%2Bsin%28x%2F2%29%2B1+=+0\"$
\n" ); document.write( " $\"2sin%5E2%28x%2F2%29-sin%28x%2F2%29-1+=+0\"$
\n" ); document.write( "Factor: $\"%28sin%28x%2F2%29-1%29%282sin%28x%2F2%29%2B1%29+=+0\"$
\n" ); document.write( "Split into two equations:
\n" ); document.write( " $\"sin%28x%2F2%29-1+=+0\"$ and $\"2sin%28x%2F2%29%2B1+=+0\"$
\n" ); document.write( "For $\"sin%28x%2F2%29-1+=+0\"$ :
\n" ); document.write( " $\"sin%28x%2F2%29+=+1\"$
\n" ); document.write( "Take the inverse Sin:
\n" ); document.write( " $\"x%2F2+=+pi%2F2\"$
\n" ); document.write( " $\"x+=+pi\"$ \r
\n" ); document.write( "\n" ); document.write( "For $\"2sin%28x%2F2%29%2B1+=+0\"$ :
\n" ); document.write( " $\"2sin%28x%2F2%29+=+-1\"$
\n" ); document.write( " $\"sin%28x%2F2%29+=+-1%2F2\"$
\n" ); document.write( "Take the inverse Sin:
\n" ); document.write( " $\"x%2F2+=+7pi%2F6\"$
\n" ); document.write( " $\"x+=+7pi%2F3\"$ , but this is not within the restriction 0 to $\"2pi\"$
\n" ); document.write( "OR
\n" ); document.write( " $\"x%2F2+=+11pi%2F6\"$
\n" ); document.write( " $\"x+=+11pi%2F3\"$ , but this is not within the restriction 0 to $\"2pi\"$ \r
\n" ); document.write( "\n" ); document.write( "So the only value of x within the restriction 0 to $\"2pi\"$ is $\"pi\"$ \n" ); document.write( "

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