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

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SUMMARY

The equation (cos2x)/(sin3x-sinx) = 1 can be solved by reducing it to a cubic equation. The key solution identified is sinx = 1/2, leading to the general solutions x = (π/6) + 2πk and x = (5π/6) + 2πk, where k is any integer. The discussion highlights the challenges faced in factoring the quadratic equation derived from the original problem, particularly when encountering negative values under the radical, which renders the equation undefined.

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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!
 

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