Katie's question at Yahoo Answers: trigonometric equation

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SUMMARY

The discussion revolves around solving the trigonometric equation \( \frac{1+\sin x}{\cos x} + \frac{\cos x}{1+\sin x} = 4 \) for \( 0 < x < 2\pi \). The correct solutions derived from the equation are \( x = \frac{\pi}{3} \) and \( x = \frac{5\pi}{3} \). The participants emphasize the importance of using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \) to simplify the equation effectively. Additionally, they discuss the significance of avoiding unnecessary complications by directly using \( \cos x \) instead of manipulating square roots.

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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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