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I am fairly good at solving trig equations yet this one equation has me stumped. I've been trying very hard but was unable to solve it. Can anyone help please? Thank you.

$$\frac{\cos x}{1+\sin x} + \frac{1+\sin x}{\cos x} = 2$$

solve for $x$ in the range of $[-2\pi, 2\pi]$

I do know we have to do a difference of squares, yet after that, I don't know what to do and I get lost.

Thank you.

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  • $\begingroup$ $$\frac{\cos x}{1+\sin x}=\tan\left(\frac{\pi}{4}-\frac{x}{2}\right)$$ might be helpful. $\endgroup$ – J. M. is a poor mathematician May 1 '12 at 18:01
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    $\begingroup$ For that matter: $u+\frac1{u}=2$ has only one solution. That can be very helpful here. $\endgroup$ – J. M. is a poor mathematician May 1 '12 at 18:05
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HINT: $$\begin{align*} \frac{\cos x}{1+\sin x} + \frac{1+\sin x}{\cos x}&=\frac{\cos^2 x+(1+\sin x)^2}{\cos x(1+\sin x)}\\ &=\frac{\cos^2 x+\sin^2x+1+2\sin x}{\cos x(1+\sin x)}\;; \end{align*}$$

now use a familiar trig identity and find something to cancel.

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  • $\begingroup$ thank you! I can't believe I didn't try this! $\endgroup$ – Outlier May 1 '12 at 18:26
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Hint: if you put the two fractions over a common denominator you get a nice cancellation.

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    $\begingroup$ I should have thought of that! +1 $\endgroup$ – Outlier May 1 '12 at 18:26
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You have a fraction and its reciprocal adding to $2$: $\lambda + 1/\lambda=2$. Only solution is $\lambda=1$, as @J.M. has noted. So $\cos x = 1+\sin x$. Draw the graphs of the two sides of the equation and see that there are five intersections in the $x$-interval $[-2\pi,2\pi]$, but the two at $x=-\pi/2$ and $x=3\pi/2$ are no good, ’cause there you get $0/0$ occurring in the original problem.

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$$ \begin {align*} & 2=\frac{\cos x}{1+\sin x} + \frac{1+\sin x}{\cos x}\\ & =\frac{\cos^2 x+(1+\sin x)^2}{\cos x(1+\sin x)}\\ & =\frac{\cos^2 x+\sin^2x+1+2\sin x}{\cos x(1+\sin x)}\\ & =\frac{2+2 \sin x}{\cos x(1+\sin x)} \, \\ & =\frac{2( 1+\sin x) }{\cos x(1+\sin x)}\;\\ & =\frac{2}{\cos x}\; \end{align*}$$ $$ \rightarrow \cos x = 1, x = 2 k \pi $$

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  • $\begingroup$ This does not seem to add much to the method of Brian M. Scott. Can you edit this to add more value? $\endgroup$ – Rory Daulton Feb 28 '16 at 12:34
  • $\begingroup$ cross multiplication simplification, old favorite method. $\endgroup$ – Narasimham Feb 28 '16 at 17:23
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After simplification, with condition $1 + \sin x \ne 0$ (or $x \ne 3\pi/2$), the given equation becomes:

$$\frac{\cos x}{(1 + \sin x)} + \frac{(1+ \sin x)}{\cos x} = \frac{2}{\cos x} = 2\;,$$

which implies that

$$\cos x = \frac22 = 1\;.$$

Calculators and the Trig Unit Circle give as answers:

$x = 0$ (or $0$ deg.) and $x = 2\pi$ (or $360$ deg.)

Let's check: when $x = 0$ or $x = 2\pi$, we have $\cos x = 1$ and $\sin x = 0$, then:

$1/1 + 1/1 = 2$ (True).

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    $\begingroup$ would be nice if you showed all the steps, and used some latex. $\endgroup$ – yiyi Sep 10 '12 at 6:00
  • $\begingroup$ Worth noting that $x=-2\pi$ also works. $\endgroup$ – Cameron Buie Sep 22 '12 at 3:30
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As $\cos x\cdot\cos x=(1-\sin x)(1+\sin x)$

$$\dfrac{\cos x}{1+\sin x}=\dfrac{1-\sin x}{\cos x}$$

So, the given relation reduces to $$\dfrac2{\cos x}=2\iff\cos x=1$$

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