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

$\begingroup$

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.

$\endgroup$ 2

6 Answers

$\begingroup$

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.

$\endgroup$ 1 $\begingroup$

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

$\endgroup$ 1 $\begingroup$

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.

$\endgroup$ $\begingroup$

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

$\endgroup$ 2 $\begingroup$

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).

$\endgroup$ 2 $\begingroup$

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 $$\dfrac2{\cos x}=2\iff\cos x=1$$

$\endgroup$

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy