document.write( "Question 1006604: what would be the solutions to (sin2x+cos2x)^2=1?
\n" ); document.write( "I know that this can be expanded to (sin2x+cos2x) (sin2x+cos2x)=1, which can be multiplied to be sin^2(4x)+2sin(2x)cos(2x)+cos^2(4x)=1. Or am I missing something here? I'm not sure what to do after this. Please help! \n" ); document.write( "

Algebra.Com's Answer #622741 by ikleyn(52814) $\"\"$ $\"About$

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\n" ); document.write( "What would be the solutions to (sin2x + cos2x)^2 = 1?
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document.write( "Start with \r\n" );
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document.write( " =  +  +  = 1 + 2*sin(2x)*cos(2x).\r\n" );
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document.write( "Therefore, your equation takes the form\r\n" );
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document.write( "1 + 2*sin(2x)*cos(2x) = 1,   or\r\n" );
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document.write( "2*sin(2x)*cos(2x) = 0.   (1)\r\n" );
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document.write( "Now, do you know this formula sin(2alpha) = 2*sin(alpha)*cos(alpha) ?\r\n" );
document.write( "(See the lesson Trigonometric functions of multiply argument in this site).\r\n" );
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document.write( "It reduces the equation (1) to\r\n" );
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document.write( "sin(4x) = 0.\r\n" );
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document.write( "The solution is x = , k = 0, +/-1, +/-2, . . . \r\n" );
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document.write( "It is the solution of your original equation.\r\n" );
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