Solving Trigonometric Equations

[frozonecom]

Homework Statement

2sin\Theta-\sqrt{2}=0

Homework Equations

The Attempt at a Solution

2sin\Theta=\sqrt{2}

\frac{2sin\Theta}{2}=\frac{\sqrt{2}}{2}

So here's my problem. Should it be positive or negative? Or just positive? Since the square root of any positive number can be positive or negative, right? Is there a possibility that the answer can be both positive and negative?

sin\Theta=\pm\frac{\sqrt{2}}{2}

if so, \Theta=45 degrees or 135 degrees ---- for positive value of sin\Theta

\Theta=-45 degrees or 315 degrees or 225 degrees ---- for negative value of sin\Theta

So, I'm confused if:

2sin\Theta-\sqrt{2}=0

2sin\Theta=\sqrt{2}

\frac{2sin\Theta}{2}=\frac{\sqrt{2}}{2}

Can be:


2sin\Theta-(\pm\sqrt{2})=0

2sin\Theta=(\pm\sqrt{2})

\frac{2sin\Theta}{2}=(\pm\frac{\sqrt{2}}{2})

So can it be?
Is it mathematically correct to think of it that way? Sorry if this is such a stu*id question. :)
Hope you can help me! :)

 

[danago]

\sqrt{2} is positive and separate from its negative counterpart, -\sqrt{2}. If the question did not say ± \sqrt{2}, then you can safely just worry about the positive value.

If, on the other hand, the question was 4sin^2 \Theta-2=0, then when you take the square root you must be sure to include the ±.

It is sort of like saying x=\sqrt{25} is equal to positive 5 only, however if i said that x^2=25, then x could be +5 or -5.

 

[frozonecom]

Thank you for clearing this out for me!
Sadly that was on my quiz and I totally included the negative root as a solution for the equation. :(

At least now I know it. Thanks for answering! :)