Katie's question at Yahoo Answers: trigonometric equation

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Discussion Overview

The discussion revolves around solving the trigonometric equation \(1+\frac{\sin x}{\cos x} + \frac{\cos x}{1+\sin x} = 4\) for \(0 < x < 2\pi\). Participants explore different approaches to simplify and solve the equation, sharing their reasoning and results.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a series of algebraic manipulations leading to the conclusion that \(\sin x = \pm\frac{\sqrt{3}}{2}\), resulting in \(x = \frac{\pi}{3}\) and \(x = \frac{4\pi}{3}\), but later acknowledges a possible error.
  • Another participant derives a different result, concluding that \(\cos x = \frac{1}{2}\), leading to \(x = \frac{\pi}{3}\) and \(x = \frac{5\pi}{3}\), and provides a detailed step-by-step explanation of their approach.
  • A third participant admits to a mistake in their earlier calculations and expresses confusion about the results, indicating a need for clarification on the mathematical steps involved.
  • There is a light-hearted exchange regarding the points awarded by the original poster, with participants joking about the scoring system.

Areas of Agreement / Disagreement

Participants present differing results for the values of \(x\) that satisfy the equation, indicating that there is no consensus on the solution. The discussion remains unresolved regarding which approach is correct.

Contextual Notes

Some participants note the importance of avoiding assumptions about the signs of roots and the implications of multiplying through by expressions that may equal zero.

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Follow the steps:

\dfrac{1+\sin x}{\cos x}+\dfrac{\cos x}{1+\sin x}=4\\<br /> \dfrac{1+\sin^2x+2\sin x+\cos^2x}{\cos x(1+\sin x)}=4\\<br /> 1+\sin^2x+2\sin x+\cos^2x=4\cos x+4\sin x\cos x

Using \cos^2x=1-\sin^2x and simplifying

2(1+\sin x)=4(1+\sin x)\sqrt{1-\sin^2x}

But 1+\sin x\neq 0 (because appears in a denominator of the initial equation), so

\sqrt{1-\sin^2x}=\frac{1}{2}.

Taking squares we get \sin x=\pm\sqrt{3}/2. As 0&lt;x&lt;2\pi, we get (Edited: the following is wrong, look at the next post) x=\dfrac{\pi}{3},\;x=\dfrac{4\pi}{3}
 
Last edited:
I get a different result:

We are given:

$\displaystyle \frac{1+\sin(x)}{\cos(x)}+\frac{\cos(x)}{1+\sin(x)}=4$ where $\displaystyle 0<x<2\pi$

Multiply through by $\displaystyle (1+\sin(x))\cos(x)$:

$\displaystyle (1+\sin(x))^2+\cos^2(x)=4(1+\sin(x))\cos(x)$

$\displaystyle 1+2\sin(x)+\sin^2(x)+\cos^2(x)=4(1+\sin(x))\cos(x)$

Using the Pythagorean identity $\displaystyle \sin^2(x)+\cos^2(x)=1$, we have:

$\displaystyle 2+2\sin(x)=4(1+\sin(x))\cos(x)$

$\displaystyle 2(1+\sin(x))=4(1+\sin(x))\cos(x)$

Since we have $\displaystyle 1+\sin(x)\ne0$, this reduces to:

$\displaystyle 2=4\cos(x)$

$\displaystyle \cos(x)=\frac{1}{2}$ and so:

$\displaystyle x=\frac{\pi}{3},\,\frac{5\pi}{3}$
 
MarkFL said:
$\displaystyle \cos(x)=\frac{1}{2}$ and so: $\displaystyle x=\frac{\pi}{3},\,\frac{5\pi}{3}$

You are right. My silly mistake: $2\pi -\frac{\pi}{3}=\frac{4\pi}{3}$ (?). Why?. I need a mathematical psychiatrist. Besides, is better to write $\cos x$ as you did instead of $\sqrt{1-\sin^2x}$, so we avoid to analyze the double sign of the root.
 
Last edited:
Hey, I feel somewhat guilty as Katie awarded me 10 points for posting the link to your reply...If I could transfer them to your account I would! (Smile)
 
MarkFL said:
Hey, I feel somewhat guilty as Katie awarded me 10 points for posting the link to your reply...If I could transfer them to your account I would! (Smile)

No wonder. According to my solution surely Katie awarded me -10 points. :)
 

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