document.write( "Question 825697: I need help with this problem... Solve the equation sqrt(2) sinx=1 on the interval [0,2π). Write your answer in terms of π. \n" ); document.write( "
Algebra.Com's Answer #497518 by jim_thompson5910(35256)\"\" \"About 
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I need help with this problem... Solve the equation sqrt(2) sinx=1 on the interval [0,2π). Write your answer in terms of π.\r
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\n" ); document.write( "\n" ); document.write( "sqrt(2)*sinx=1\r
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\n" ); document.write( "\n" ); document.write( "sinx = 1/sqrt(2)\r
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\n" ); document.write( "\n" ); document.write( "sinx = sqrt(2)/2 ... rationalize the denominator\r
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\n" ); document.write( "\n" ); document.write( "x = arcsin(sqrt(2)/2) ... take the arcsine of both sides\r
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\n" ); document.write( "\n" ); document.write( "x = pi/4 or x = 3pi/4 ... use the unit circle (link: http://www.regentsprep.org/Regents/math/algtrig/ATT5/600px-Unit_circle_angles_svg.jpg)\r
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\n" ); document.write( "\n" ); document.write( "The two solutions are \"x+=+pi%2F4\" or \"x+=+3pi%2F4\"
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