# Katie's question at Yahoo! Answers: trigonometric equation

MHB Math Helper

#### Fernando Revilla

##### Well-known member
MHB Math Helper

$$\dfrac{1+\sin x}{\cos x}+\dfrac{\cos x}{1+\sin x}=4\\ \dfrac{1+\sin^2x+2\sin x+\cos^2x}{\cos x(1+\sin x)}=4\\ 1+\sin^2x+2\sin x+\cos^2x=4\cos x+4\sin x\cos x$$

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 $$0<x<2\pi$$, 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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#### MarkFL

Staff member
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:

$\displaystyle x=\frac{\pi}{3},\,\frac{5\pi}{3}$

#### Fernando Revilla

##### Well-known member
MHB Math Helper
$\displaystyle \cos(x)=\frac{1}{2}$ 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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