# Thread: [SOLVED] solve trigonometric equation

1. ## [SOLVED] solve trigonometric equation

Can someone assist in how I should solve the following equation?

$2sinx + cscx = 0$

I know that can be converted to

$2sinx + \frac{1}{sinx} = 0$

but I'm stumped as to where to go from there. Any pointers would be appreciated.

*edit*
Hah, I hadn't even looked in the back of the book prior to posting this.. it states there's no solution. Is that because there's basically no way to solve the above equation?

2. So? Prove it. It's a one-step demonstration.

3. so is it just

$sinx = -\frac{1}{2sinx}$ as the proof, since that's obviously invalid?

*edit* left off a negative

4. Why? Explain reasoning.

"Obvious" is almost never as obvious as you think.

I made the "obviously" mistake once. My professor of many years ago wrote on my proof, "To whom?". I never assumed that again.

5. Fair point on obviousness.

I'm not really sure how I would prove that it's not an equality. I can fairly easily plug in numbers... if I throw in $\frac{pi}{2}$ on both sides, I get $1 = -\frac{1}{2}$, but I don't know how to prove the inequality algebraically.

6. Originally Posted by satis
Can someone assist in how I should solve the following equation?

$2sinx + cscx = 0$

I know that can be converted to

$2sinx + \frac{1}{sinx} = 0$

but I'm stumped as to where to go from there. Any pointers would be appreciated.

*edit*
Hah, I hadn't even looked in the back of the book prior to posting this.. it states there's no solution. Is that because there's basically no way to solve the above equation?
$2\sin{x} + \csc{x} = 0$

$2\sin{x} + \frac{1}{\sin{x}} = 0$.

Multiply both sides by $\sin{x}$ to get

$2\sin^2{x} + 1 = 0$

$2\sin^2{x} = -1$

$\sin^2{x} = -\frac{1}{2}$

I think you can see from here that no real solution exists.

7. ahh, I see, I went awry when I failed to keep the variable on one side of the equation. Thanks for the assistance.

8. Yours was fine. There are many types of statements.

Prove It brought it algebraically to a pretty obvious place, but yours was there already.

Sibstituting A = sin(x)

Your structure $2\cdot A = -\frac{1}{A}$

0) The '2' is of little consequence in this argument. I'll mention only sign, not magnitude.

1) What number's recipriocal can have the opposite sign? Only zero (0) could possibly have this property, since zero (0) DOES have this property if we don't mention the reciprocal. We've narrowed it down to one possible suspect.

2) 'A' cannot be zero on the right hand side. Of course, zero (0) doesn't have a reciprocal. We just eliminated our ONLY suspect.

3) Done.

This IS a little ugly, but any valid argument should be acceptable.