MHB Katie's question at Yahoo Answers: trigonometric equation

AI Thread Summary
The discussion centers around solving the trigonometric equation 1+sin(x)/cos(x) + cos(x)/(1+sin(x)) = 4 for 0 < x < 2π. The solution involves multiplying through by (1+sin(x))cos(x) and simplifying using the Pythagorean identity. This leads to the equation 2 = 4cos(x), resulting in cos(x) = 1/2. The valid solutions for x in the specified range are x = π/3 and x = 5π/3. The participants also reflect on the solution process and share a light-hearted exchange about the points awarded for responses.
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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}
 
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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.
 
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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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