document.write( "Question 234142: SOLVE EACH EQUATION, WHERE 0 DEGREES (< OR = TO) X < 360 DEGREES. ROUND APPROXIMATE SOLUTIONS TO THE NEAREST TENTH OF A DEGREE 3 SIN X - 5 + CSC X = 0 \n" ); document.write( "
Algebra.Com's Answer #172722 by Edwin McCravy(20056)\"\" \"About 
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SOLVE EACH EQUATION, WHERE 0 DEGREES (< OR = TO) X < 360 DEGREES. ROUND APPROXIMATE SOLUTIONS TO THE NEAREST TENTH OF A DEGREE 3 SIN X - 5 + CSC X = 0
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document.write( "\"3%2ASin%28X%29+-+5+%2B+Csc%28X%29+=+0\"\r\n" );
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document.write( "Change \"Csc%28X%29\" to \"1%2FSin%28X%29\"\r\n" );
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document.write( "\"3%2ASin%28X%29+-+5+%2B+1%2FSin%28X%29+=+0\"\r\n" );
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document.write( "Multiply through by \"Sin%28X%29\" to clear of fractions:\r\n" );
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document.write( "\"3%2ASin%5E2%28X%29+-+5%2ASin%28X%29+%2B+1+=+0\"\r\n" );
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document.write( "\"3%2ASin%5E2%28X%29+-+5%2ASin%28X%29+%2B+1+=+0\"\r\n" );
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document.write( "Use the quadratic formula:\r\n" );
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document.write( "where \"x=Sin%28X%29\", \"a+=+3\", \"b=-5\", and \"c=1\"\r\n" );
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document.write( "\"x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\" \r\n" );
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document.write( "\"Sin%28X%29+=+%285+%2B-+sqrt%2825-12+%29%29%2F6+\"\r\n" );
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document.write( "\"Sin%28X%29+=+%285+%2B-+sqrt%2813%29%29%2F6+\"\r\n" );
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document.write( "Using the +\r\n" );
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document.write( "\"Sin%28X%29+=+%285+%2B+sqrt%2813%29%29%2F6+=+1.434258546\"\r\n" );
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document.write( "That's impossible because all sines are between -1 and +1.\r\n" );
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document.write( "Using the -\r\n" );
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document.write( "\"Sin%28X%29+=+%285+-+sqrt%2813%29%29%2F6+=+0.2324081208\"\r\n" );
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document.write( "That is a positive number and is between -1 and +1,\r\n" );
document.write( "so that will give us two solutions, since the sine\r\n" );
document.write( "is positive in quadrant I and quadrant II.\r\n" );
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document.write( "To find the quadrant I solution, we find the\r\n" );
document.write( "inverse sine of 0.2324081208 which is\r\n" );
document.write( "13.56526739° which rounds to 13.6° to the nearest\r\n" );
document.write( "tenth of a degree.\r\n" );
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document.write( "That's the reference angle, and in Quadrant I, the\r\n" );
document.write( "solution IS the reference angle, 13.6°\r\n" );
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document.write( "To find the quadrant II solution, we subtract the\r\n" );
document.write( "reference angle, 13.6° from 180° and get 166.4°\r\n" );
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document.write( "So the two solutions are X = 13.6° and X = 166.4°.\r\n" );
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document.write( "Edwin
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