SOLUTION: solve 2sin x + sqrt 3<0 for 0<= x < 2pi a)4pi/3 < x < 5pi/3 b)2pi/3 < x < 4pi/3 c)7pi/6 < x < 11pi/6 d)5pi/6 < x < 7pi/6

Algebra ->  Trigonometry-basics -> SOLUTION: solve 2sin x + sqrt 3<0 for 0<= x < 2pi a)4pi/3 < x < 5pi/3 b)2pi/3 < x < 4pi/3 c)7pi/6 < x < 11pi/6 d)5pi/6 < x < 7pi/6      Log On


   



Question 215576: solve 2sin x + sqrt 3<0 for 0<= x < 2pi
a)4pi/3 < x < 5pi/3
b)2pi/3 < x < 4pi/3
c)7pi/6 < x < 11pi/6
d)5pi/6 < x < 7pi/6

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
solve 2sin x + sqrt 3<0 for 0<= x < 2pi
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sin(x) < -sqrt(3)/2
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sin is negative in the III and IV quadrants.
The reference angle for sin(x)) = sqrt(3)/2 is pi/3
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In QIII the angle is (4/3)pi
In QIV the angle is (5/3)pi
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Answer: a
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Cheers,
Stan H.


a)4pi/3 < x < 5pi/3
b)2pi/3 < x < 4pi/3
c)7pi/6 < x < 11pi/6
d)5pi/6 < x < 7pi/6