Solving Trigonometric Equations with multiple solutions

Sadriam
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Homework Statement


Solve each equation for solutions on the interval 0 ≤ x ≤ 2∏.
Code:
2 sin x = √3

The Attempt at a Solution


Okay so I was able to solve this one:

  • 2 sin x = √3
  • sin x = (√3) / 2

So i got x = ∏/3 ; 2∏/3 ; 4∏/3 ; 5∏/3

I substituted the x's into the original problem, and ∏/3 and 2∏/3 give +(√3). The other two give -(√3). Are the one's that give negative results also solutions? If so, is this only true for (±√x), or is the sign always a significant factor?
 
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No, if sin(x)= -\sqrt{3}/2 then 2sin(x)= -\sqrt{3} which is NOT the original equation. Those values of x do NOT satisfy the equation and are NOT solutions. In particular, \sqrt{a} is NOT the same as either -\sqrt{a} or \pm\sqrt{a}.
 
Thank you for clarifying (:
 
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