document.write( "Question 1036995: Solve trig equation on the interval [0,2pi]:
\n" ); document.write( "sin^2x-cos^2x=0 \n" ); document.write( "
Algebra.Com's Answer #651695 by Alan3354(69443)\"\" \"About 
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Solve trig equation on the interval [0,2pi]:
\n" ); document.write( "sin^2x-cos^2x=0
\n" ); document.write( "---
\n" ); document.write( "sin^2 - (1-sin^2) = 0
\n" ); document.write( "2sin^2 - 1 = 0
\n" ); document.write( "sin = ħsqrt(1/2)
\n" ); document.write( "x = pi/4, 3pi/4, 5pi/4, 7pi/4\r
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\n" ); document.write( "Or,
\n" ); document.write( "sin^2x-cos^2x=0
\n" ); document.write( "(sin + cos)*(sin - cos) = 0
\n" ); document.write( "sin = cos
\n" ); document.write( "sin = -cos
\n" ); document.write( "Same as above
\n" ); document.write( " \n" ); document.write( "
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