document.write( "Question 1054226: Solve for 0≤theta≤360, 2cos^2 theta +5cos theta-3=0 \n" ); document.write( "

Algebra.Com's Answer #669428 by ikleyn(52809) $\"\"$ $\"About$

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\n" ); document.write( "Solve for 0≤theta≤360, 2cos^2 theta +5cos theta-3=0
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document.write( " = 0,\r\n" );
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document.write( "Introduce new variable x = . Then your equation takes the form\r\n" );
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document.write( " = 0.\r\n" );
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document.write( "Apply the quadratic formula:\r\n" );
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document.write( " =  = .\r\n" );
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document.write( " = ,   = -3.\r\n" );
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document.write( "The second root is not acceptable, since cosine must be <= 1 by modulus.\r\n" );
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document.write( "The only solution is  = ,\r\n" );
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document.write( "which gives  = 60 degs and/or   = 300 degs.\r\n" );
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document.write( "Answer.   = 60 degs  and/or   = 300 degs.\r\n" );
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