document.write( "Question 1036282: Find all solutions of the equation in the interval [0,2pi).\r
\n" ); document.write( "\n" ); document.write( "sin^2x=2-2cosx\r
\n" ); document.write( "\n" ); document.write( "Write your answer in radians in terms of pi.
\n" ); document.write( "If there is more than one solution, separate them with commas.
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Algebra.Com's Answer #650942 by Cromlix(4381)\"\" \"About 
You can put this solution on YOUR website!
Hi there,
\n" ); document.write( "sin^2x = 2 - 2cosx
\n" ); document.write( "Using cos^2x + sin^2x = 1
\n" ); document.write( "so, sin^2x = 1 - cos^2x
\n" ); document.write( "1 - cos^2x = 2 - 2cosx
\n" ); document.write( "Rearrange.
\n" ); document.write( "cos^2x - 2cosx + 1 = 0
\n" ); document.write( "(cosx - 1)(cosx - 1) = 0
\n" ); document.write( "cosx - 1 = 0
\n" ); document.write( "cosx = 1
\n" ); document.write( "x = 0, 2π
\n" ); document.write( "Hope this helps :-)
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