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

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Homework Help Overview

The discussion revolves around solving the equation (cos2x)/(sin3x-sinx) = 1, which involves trigonometric functions and potentially leads to a quadratic equation. Participants express challenges in finding solutions and understanding the implications of their calculations.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss their attempts to manipulate the equation, with one mentioning reaching a quadratic form that does not factor. Another participant raises concerns about encountering a negative value under the radical, suggesting the equation may be undefined. There is also a mention of deriving specific values for sinx and their corresponding x values.

Discussion Status

The discussion is ongoing, with participants sharing their findings and questioning the validity of their approaches. One participant has suggested a potential solution, while another has pointed out the lack of detailed work presented, indicating a need for further clarification and exploration of the problem.

Contextual Notes

There is a mention of difficulties in factoring a quadratic equation and concerns about undefined values in the context of the problem. Participants are also navigating the representation of mathematical symbols in their posts.

xtpacygax
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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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