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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.

  • 4
    For that matter: $u+\frac1{u}=2$ has only *one* solution. That can be very helpful here.2012-05-01

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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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    thank you! I can't believe I didn't try this!2012-05-01
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Hint: if you put the two fractions over a common denominator you get a nice cancellation.

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    I should have thought of that! +12012-05-01
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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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    cross multiplication simplification, old favorite method.2016-02-28
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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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    Worth noting that $x=-2\pi$ also works.2012-09-22
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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 $\dfrac2{\cos x}=2\iff\cos x=1$