Trigonometry equation

Hi trivun. That equation has no Real solutions. Did you copy it correctly?

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trivun said:
actually the question is how many real solutions are there.
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Why not tell us the exercise up front? (Please see the Read Before Posting announcement.)

We don't need much calculation, to answer that question. Think about the interesting values of sin(x), on a graph over one period:

0 \(\;\) 1 \(\;\) 0 \(\;\) -1 \(\;\) 0

If we add the squares of those to the reciprocal of the squares of those, what do we get? In between the values above, we can interpolate because squares are never negative.

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Let [MATH]u = sin(x) \text { and } sin^2(x) + \dfrac{1}{sin^2(x)} = sin(x) \implies[/MATH]
[MATH]0 < |\ u \ | \le 1 \text { and } u^2 + \dfrac{1}{u^2} = u \implies[/MATH]
[MATH]u^4 - u^3 + 1 = 0.[/MATH]
[MATH]0 < u^4 \le 1 \implies 1 < u^4 + 1 \le 2.[/MATH]
[MATH]- 1 \le - u^3 \le 1. [/MATH]
From that we conclude what?
 
JeffM said:
Let [MATH]u = sin(x) \text { and } sin^2(x) + \dfrac{1}{sin^2(x)} = sin(x) \implies[/MATH]
[MATH]0 < |\ u \ | \le 1 \text { and } u^2 + \dfrac{1}{u^2} = u \implies[/MATH]
[MATH]u^4 - u^3 + 1 = 0.[/MATH]
[MATH]0 < u^4 \le 1 \implies 1 < u^4 + 1 \le 2.[/MATH]
[MATH]- 1 \le - u^3 \le 1. [/MATH]
From that we conclude what?
Click to expand...
Thank you.
 
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