document.write( "Question 642735: solve for x over the interval -2pi,2pi: 2sin(x+(pi/4))=1. Can you also show how you got the answer \n" ); document.write( "

Algebra.Com's Answer #404209 by lwsshak3(11628) $\"\"$ $\"About$

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solve for x over the interval -2pi,2pi: 2sin(x+(pi/4))=1
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\n" ); document.write( " -2pi,2pi is not a correct expression for an interval
\n" ); document.write( "I will assume an interval of [0,2π)
\n" ); document.write( "2sin(x+π/4))=1
\n" ); document.write( "sin(x+π/4))=1/2
\n" ); document.write( "x+π/4=π/6 and 5π/6 (in quadrants I and II where sin>0)
\n" ); document.write( "x=π/6-π/4=2π/12-3π/12=-π/12
\n" ); document.write( "or
\n" ); document.write( "x=5π/6-π/4=10π/12-3π/12=7π/12
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