document.write( "Question 215576: solve 2sin x + sqrt 3<0 for 0<= x < 2pi\r
\n" ); document.write( "\n" ); document.write( "a)4pi/3 < x < 5pi/3
\n" ); document.write( "b)2pi/3 < x < 4pi/3
\n" ); document.write( "c)7pi/6 < x < 11pi/6
\n" ); document.write( "d)5pi/6 < x < 7pi/6
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Algebra.Com's Answer #162881 by stanbon(75887)\"\" \"About 
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solve 2sin x + sqrt 3<0 for 0<= x < 2pi
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\n" ); document.write( "sin(x) < -sqrt(3)/2
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\n" ); document.write( "sin is negative in the III and IV quadrants.
\n" ); document.write( "The reference angle for sin(x)) = sqrt(3)/2 is pi/3
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\n" ); document.write( "In QIII the angle is (4/3)pi
\n" ); document.write( "In QIV the angle is (5/3)pi
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\n" ); document.write( "Answer: a
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.\r
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\n" ); document.write( "\n" ); document.write( "a)4pi/3 < x < 5pi/3
\n" ); document.write( "b)2pi/3 < x < 4pi/3
\n" ); document.write( "c)7pi/6 < x < 11pi/6
\n" ); document.write( "d)5pi/6 < x < 7pi/6
\n" ); document.write( "
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