Find all the solutions in the interval [0, 2pi) : csc x + 2 = 0.?

Question

a. (pi/3) , (2pi/3)
b. (pi/6) , (5pi/6)
c. (4pi/3) , (5pi/3)
d. (7pi/6) , (11pi/6)
e. none of these

Answer 1

csc(x) + 2 = 0

1/sin(x) + 2

2sin(x) = -1

sin(x) = -1/2

The sin curve is negative from the interval of [pi, 2pi] and sin gives a value of 1/2 only when the reference angle is pi/6

So your two answers are 7pi/6 and 11pi/6 - d

Answer 2

csc x + 2 = 0 is the same as saying 1/sin x = -2, or sin x = -1/2. In order for an angle in standard position to have a sine of -1/2, the reference angle has to be 30 degrees = pi/6 radians, and the terminal side must be in quadrant III or quadrant IV. The two angles satisfying this are 7pi/6 and 11pi/6, so the answer is (d).