SOLUTION: Can you please help me. Please! Solve 4sin(square)A + 1 = sin(square)A + 2 to the nearest ten minutes or nearest tenth of a degree in the interval ( 0ș less than or equal to

Algebra ->  Trigonometry-basics -> SOLUTION: Can you please help me. Please! Solve 4sin(square)A + 1 = sin(square)A + 2 to the nearest ten minutes or nearest tenth of a degree in the interval ( 0ș less than or equal to       Log On


   

Question 373951: Can you please help me. Please!
Solve 4sin(square)A + 1 = sin(square)A + 2 to the nearest ten minutes or nearest tenth of a degree in the interval ( 0ș less than or equal to A less than 360ș)
Thank you very much!
Answer by Edwin McCravy(20086) About Me  (Show Source):
You can put this solution on YOUR website!
 4sinČA + 1 = sinČA + 2
 -sinČA      -sinČA
-----------------------
 3sinČA + 1 =         2
        - 1          -1
-----------------------
 3sinČA     =         1

      sinČA = 1%2F3
     sin(A) = %22%22%2B-sqrt%281%2F3%29

Use the inverse sine function on the calculator to
find the reference angle for A which I call "ref(A)"

     ref(A) = 35.26438968°

Since the sign is either + or -, we must include all angles
with reference angle 35.3° between 0° and 360°

The reference angle is the first quadrant answer which to 
the nearest tenth of a degree is 

35.3°

the second quadrant angle between 0° and 360° with reference 
angle 35.3° is gotten by subtracting 35.3° from 180°.

180°-35.3° = 144.7°

the third quadrant angle between 0° and 360° with reference 
angle 35.3° is gotten by adding 35.3° to 180°.

180°+35.3° = 215.3°

the fourth quadrant angle between 0° and 360° with reference 
angle 35.3° is gotten by subtracting 35.3° from 360°.

360°-35.3° = 324.7°

Those are the four solutions.

Edwin