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## Homework Statement

[itex]2sin\Theta-\sqrt{2}=0[/itex]

## Homework Equations

## The Attempt at a Solution

[itex]2sin\Theta=\sqrt{2}[/itex]

[itex]\frac{2sin\Theta}{2}=\frac{\sqrt{2}}{2}[/itex]

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?

[itex]sin\Theta=\pm\frac{\sqrt{2}}{2}[/itex]

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

[itex]\Theta=-45 degrees[/itex] [itex] or [/itex] [itex]315 degrees[/itex] [itex] or[/itex] [itex] 225 degrees[/itex] ---- for negative value of [itex]sin\Theta[/itex]

So, I'm confused if:

[itex]2sin\Theta-\sqrt{2}=0[/itex]

[itex]2sin\Theta=\sqrt{2}[/itex]

[itex]\frac{2sin\Theta}{2}=\frac{\sqrt{2}}{2}[/itex]

Can be:

[itex]2sin\Theta-(\pm\sqrt{2})=0[/itex]

[itex]2sin\Theta=(\pm\sqrt{2})[/itex]

[itex]\frac{2sin\Theta}{2}=(\pm\frac{\sqrt{2}}{2})[/itex]

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! :)