MHB Katie's question at Yahoo Answers: trigonometric equation

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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. :)
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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