document.write( "Question 553772: Solve ( 2sin^2x - cos^2x - 2 = 0 ) for ( 0° ≤ θ < 360° ). \n" ); document.write( "

Algebra.Com's Answer #361146 by lwsshak3(11628) $\"\"$ $\"About$

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Solve ( 2sin^2x - cos^2x - 2 = 0 ) for ( 0° ≤ θ < 360° )
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\n" ); document.write( "2sin^2x-cos^2x-2=0
\n" ); document.write( "2sin^2x-(1-sin^2x)-2=0
\n" ); document.write( "2sin^2x-1+sin^2x-2=0
\n" ); document.write( "3sin^2x-3=0
\n" ); document.write( "3sin^2x=3
\n" ); document.write( "sin^2x=1
\n" ); document.write( "sinx=±1
\n" ); document.write( "x=90º and 270º \n" ); document.write( "

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