document.write( "Question 594974: Solve algebraically for all values of x in the interval 0≤ x < 360.2 sin2 x− 4 sin x = cos2 x − 2
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Algebra.Com's Answer #377000 by lwsshak3(11628)\"\" \"About 
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Solve algebraically for all values of x in the interval 0≤ x < 360.2 sin2 x− 4 sin x = cos2 x − 2
\n" ); document.write( "Express your answers to the nearest degree.
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\n" ); document.write( "2 sin2 x− 4 sin x = cos2 x − 2
\n" ); document.write( "2sin^2-4sin=cos^2-2
\n" ); document.write( "2sin^2-4sin=1-sin^2-2
\n" ); document.write( "3sin^2-4sin+1=0
\n" ); document.write( "(3sinx-1)(sinx-1)=0
\n" ); document.write( "..
\n" ); document.write( "3sinx+1=0
\n" ); document.write( "sinx=-1/3
\n" ); document.write( "x≈199º and 341º (in quadrants III and IV where sin<0)
\n" ); document.write( "..
\n" ); document.write( "sinx-1=0
\n" ); document.write( "sinx=1
\n" ); document.write( "x=90º
\n" ); document.write( "ans:
\n" ); document.write( "x=90º, 199º and 341º
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