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

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.

-
$$\frac{\cos x}{1+\sin x}=\tan\left(\frac{\pi}{4}-\frac{x}{2}\right)$$ might be helpful. – J. M. May 1 '12 at 18:01
For that matter: $u+\frac1{u}=2$ has only one solution. That can be very helpful here. – J. M. May 1 '12 at 18:05

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.

-
thank you! I can't believe I didn't try this! – Outlier May 1 '12 at 18:26

Hint: if you put the two fractions over a common denominator you get a nice cancellation.

-
I should have thought of that! +1 – Outlier May 1 '12 at 18:26

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.

-

\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*} $$\rightarrow \cos x = 1, x = 2 k \pi$$

-
This does not seem to add much to the method of Brian M. Scott. Can you edit this to add more value? – Rory Daulton Feb 28 at 12:34
cross multiplication simplification, old favorite method. – Narasimham Feb 28 at 17:23
Thank you, much improved! – Rory Daulton Feb 28 at 18:00

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).

-
would be nice if you showed all the steps, and used some latex. – yiyi Sep 10 '12 at 6:00
Worth noting that $x=-2\pi$ also works. – Cameron Buie Sep 22 '12 at 3:30
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://math.stackexchange.com/questions/139508/solve-frac-cos-x1-sin-x-frac1-sin-x-cos-x-2", "fetch_time": 1464572239000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2016-22/segments/1464049282327.67/warc/CC-MAIN-20160524002122-00228-ip-10-185-217-139.ec2.internal.warc.gz", "warc_record_offset": 187839709, "warc_record_length": 20106, "token_count": 1074, "char_count": 2841, "score": 4.1875, "int_score": 4, "crawl": "CC-MAIN-2016-22", "snapshot_type": "longest", "language": "en", "language_score": 0.8613131642341614, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00029-of-00064.parquet"}