SOLUTION: find the smallest positive x(in terms of π)for which sinx=-sqrt( 3 )/2 a.x=2π/3,b.x=4π/3,c.x=5π/6,d.x=7π/6 I believe the correct answer is A.

Algebra ->  Trigonometry-basics -> SOLUTION: find the smallest positive x(in terms of π)for which sinx=-sqrt( 3 )/2 a.x=2π/3,b.x=4π/3,c.x=5π/6,d.x=7π/6 I believe the correct answer is A.      Log On


   

Question 1144204: find the smallest positive x(in terms of π)for which sinx=-sqrt( 3 )/2
a.x=2π/3,b.x=4π/3,c.x=5π/6,d.x=7π/6
I believe the correct answer is A.
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52804) About Me  (Show Source):
You can put this solution on YOUR website!
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find the smallest positive x(in terms of π)for which sin(x) = -sqrt( 3 )/2
a.x=2π/3,b.x=4π/3,c.x=5π/6,d.x=7π/6
~~~~~~~~~~~~~~~ 

The correct answer is  b. x = 4pi%2F3.


What you do believe, it does not matter, since you don't know the subject.


Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


sin%2860%29+=+sqrt%283%29%2F2

A 60 degree reference angle in any quadrant will give you a sine value of sqrt%283%29%2F2 or -sqrt%283%29%2F2.

Sine is positive in quadrants I and II; so the smallest positive angle with a sine of -sqrt%283%29%2F2 will be in quadrant III.

A 60 degree reference angle in quadrant III is 240 degrees.

Since the answer choices are in radians, 240 degrees is 4π/3 radians.

So answer b.

Note for your answer a the reference angle is the correct 60 degrees or π/3 radians; but it is in quadrant II, where the sine value is positive, not negative.