SOLUTION: 5) Solve for all values of 2 sin θ + sqrt 3=0 when 0° ≤ θ ≤ 360°. a)240°, 300° b)150°, 210° c)60°, 120° d)60°, 300° Answer: a Can you please check to

Algebra ->  Trigonometry-basics -> SOLUTION: 5) Solve for all values of 2 sin θ + sqrt 3=0 when 0° ≤ θ ≤ 360°. a)240°, 300° b)150°, 210° c)60°, 120° d)60°, 300° Answer: a Can you please check to      Log On


   

Question 388841: 5) Solve for all values of 2 sin θ + sqrt 3=0 when 0° ≤ θ ≤ 360°.
a)240°, 300° b)150°, 210° c)60°, 120° d)60°, 300°
Answer: a
Can you please check to see If did it right.
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!

2sin%28theta%29+%2B+sqrt%283%29=0 when %220%B0%22+%3C=+theta+%3C+%22360%B0%22

2sin%28theta%29+%2B+sqrt%283%29=0
2sin%28theta%29+=-sqrt%283%29
sin%28theta%29+=+-sqrt%283%29%2F2

We know that sin%28%2260%B0%22%29+=+%22%22%2Bsqrt%283%29%2F2, not -sqrt%283%29%2F2

but that tells us that the reference angle of q is 60°.

We also know that the sine is a negative number in the third and fourth
quadrants.  To find the angle whose reference angle is 60° in the 3rd
quadrant we add 180° to the reference angle 60°, and get 240°.  To find the
angle whose reference angle is 60° in the 4th quadrant we subtract the
reference angle 60° from 360° and get 300°.   

So yes you're right.

Edwin