Solving for All Values of x in (cos2x)/(sin3x-sinx) = 1

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The equation (cos2x)/(sin3x-sinx) = 1 leads to the solution sinx = 1/2, resulting in x = (pi)/6 + 2(pi)k and x = 5(pi)/6 + 2(pi)k. The original poster struggled with the quadratic form, encountering issues with undefined values under the radical. Another participant pointed out that reducing the equation to sin x yields a cubic equation with a clear root. They also noted that the remaining quadratic simplifies to 2x^2 - 1 = 0. The discussion highlights the challenges in solving trigonometric equations and the importance of showing work for clarity.
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Solve for all values of x:

(cos2x)/(sin3x-sinx) = 1

I found this problem and am having an extremely tough time solving it. I gotten down to a quadratic equation a couple of times, but it wouldn't factor. I am really stumped and really need some help. Thanks in advance.
 
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xtpacygax said:
Solve for all values of x:

(cos2x)/(sin3x-sinx) = 1

I found this problem and am having an extremely tough time solving it. I gotten down to a quadratic equation a couple of times, but it wouldn't factor. I am really stumped and really need some help. Thanks in advance.

Don't you know how to solve all quadratic equations?
 
Yes, I know how to solve quadratic equations, but I would get a negative under the radical and it would be undefined so the equation would not be possible. I think I found an answer and would like if someone could confirm this.

For the final equation I got:

sinx = 1/2

and then I got:

x = (pi)/6 + 2(pi)k
x = 5(pi)/6 + 2(pi)k

(Sorry, I don't know how to make a pi symbol)
 
You have shown really none of your work so it's impossible to say where you might have gone wrong. Reducing the orginal formula to sin x, I get a cubic equation which has an obvious root of sin x= 1/2 and the remaining quadratic is just 2x2- 1= 0!
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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