# Solving Trigonometric Equations

1. Nov 16, 2011

### frozonecom

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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! :)

2. Nov 16, 2011

### 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.

3. Nov 16, 2011

### 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! :)