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

Algebra ->  Trigonometry-basics -> 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      Log On


   

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) About Me  (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
 
3%2ASin%28X%29+-+5+%2B+Csc%28X%29+=+0

Change Csc%28X%29 to 1%2FSin%28X%29

3%2ASin%28X%29+-+5+%2B+1%2FSin%28X%29+=+0

Multiply through by Sin%28X%29 to clear of fractions:





3%2ASin%5E2%28X%29+-+5%2ASin%28X%29+%2B+1+=+0

3%2ASin%5E2%28X%29+-+5%2ASin%28X%29+%2B+1+=+0

Use the quadratic formula:

where x=Sin%28X%29, a+=+3, b=-5, and c=1

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ 



Sin%28X%29+=+%285+%2B-+sqrt%2825-12+%29%29%2F6+

Sin%28X%29+=+%285+%2B-+sqrt%2813%29%29%2F6+

Using the +

Sin%28X%29+=+%285+%2B+sqrt%2813%29%29%2F6+=+1.434258546

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

Using the -

Sin%28X%29+=+%285+-+sqrt%2813%29%29%2F6+=+0.2324081208

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