SOLUTION: 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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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
Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
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
 


Change  to 



Multiply through by  to clear of fractions:









Use the quadratic formula:

where , , , and 

 







Using the +



That's impossible because all sines are between -1 and +1.

Using the -



That is a positive number and is between -1 and +1,
so that will give us two solutions, since the sine
is positive in quadrant I and quadrant II.

To find the quadrant I solution, we find the
inverse sine of 0.2324081208 which is
13.56526739° which rounds to 13.6° to the nearest
tenth of a degree.

That's the reference angle, and in Quadrant I, the
solution IS the reference angle, 13.6°

To find the quadrant II solution, we subtract the
reference angle, 13.6° from 180° and get 166.4°

So the two solutions are X = 13.6° and X = 166.4°.

Edwin

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