Expert: Paul Klarreich - 11/14/2008

Question
How do I solve this equation given that 0 is less than or = to x < 2pie: sin 2x + sqrt3/2 = 0?

I've tried to find similar problems to workt his out, but was unsuccessful. Thanks for any help!

Answer
Questioner:   Keli
Category:  Calculus
 
Question:  How do I solve this equation given that 0 is less than or = to x < 2pie: sin 2x + sqrt3/2 = 0?

I've tried to find similar problems to workt his out, but was unsuccessful. Thanks for any help!
........................
Hi, Keli,
Look for examples with subject line Trigonometric Equations.  You'll find plenty.

For your equation:

sin(2x) + sqrt(3)/2 = 0,    0 <= x < 2pi  (a usual requirement)

sin(2x) = - sqrt(3)/2

You are supposed to (really) know the stuff about the special angles: 30,45,60- degrees, so you will know that:

1. sin 60 = sqrt(3)/2  [yes I know it says minus]

2. the solutions for 2x are in Quads 3,4, so

2x = 240 = 4pi/3 + 2 pi n
2x = 300 = 5pi/3 + 2 pi n


x = 2pi/3 +  pi n
x = 5pi/6 +  pi n

Now you can have n = 0 and n = 1 and still have  0 <= x < 2pi

x = 2pi/3   
x = 5pi/6
x = 2pi/3 +  pi = 5pi/3
x = 5pi/6 +  pi = 11pi/3