Solving Trigonometric Equations with multiple solutions

[Sadriam]

Homework Statement

Solve each equation for solutions on the interval 0 ≤ x ≤ 2∏.

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?

 

[HallsofIvy]

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}.

 

[Sadriam]

Thank you for clarifying (: