# Solve for $x$; $\tan x+\sec x=2\cos x;-\infty\lt x\lt\infty$

Solve for $x$; $\tan x+\sec x=2\cos x;-\infty\lt x\lt\infty$ $$\tan x+\sec x=2\cos x$$

$$\left(\dfrac{\sin x}{\cos x}\right)+\left(\dfrac{1}{\cos x}\right)=2\cos x$$ $$\left(\dfrac{\sin x+1}{\cos x}\right)=2\cos x$$ $$\sin x+1=2\cos^2x$$ $$2\cos^2x-\sin x+1=0$$
Edit: $$2\cos^2x=\sin x+1$$ $$2(1-\sin^2x)=\sin x+1$$ $$2\sin^2x+\sin x-1=0$$ $\sin x=a$ $$2a^2+a-1=0$$ $$(a+1)(2a-1)=0$$ $$a=-1,\dfrac{1}{2}$$ $$\arcsin(-1)=-90^\circ=-\dfrac{\pi}{2}$$ $$\arcsin\left(\dfrac{1}{2}\right)=30^\circ=\dfrac{\pi}{6}$$ $$180^\circ-30^\circ=150^\circ=\dfrac{5 \pi}{6}$$ $$x=\dfrac{\pi}{6},-\dfrac{\pi}{2},\dfrac{5 \pi}{6}$$ I actually do not know if those are the only answers considering my range is infinite:$-\infty\lt x\lt\infty$

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This is incorrect; you all of a sudden dropped the term $\frac{1}{\cos(x)}$ from one side of your equation. Remember, anything you do to one side of an equation you must do to the other. –  Zev Chonoles Jul 17 '12 at 3:29
I thought I added it properly: $$\left(\dfrac{\sin x}{\cos x}\right)+\left(\dfrac{1}{\cos x}\right)=\dfrac{\sin x}{\cos x}$$ same denominator? $$\dfrac{1}{3}+\dfrac{1}{3}=\dfrac{2}{3}$$ –  Austin Broussard Jul 17 '12 at 3:31
${\sin x\over\cos x}+{1\over\cos x}={\sin x+1\over \cos x}$. –  David Mitra Jul 17 '12 at 3:32
Okay, so how do I tackle the new problem. There are still two different functions. –  Austin Broussard Jul 17 '12 at 3:34
Write the right hand side of $\sin x+1=2\cos^2 x$ as $2(1-\sin^2 x)$. –  David Mitra Jul 17 '12 at 3:36

You correctly applied the definitions of $\tan$ and $\sec$ to go from the equation in the problem statement to $$\left(\dfrac{\sin x}{\cos x}\right)+\left(\dfrac{1}{\cos x}\right)=2\cos x.$$ However, you combined the two fractions incorrectly; in general $$\frac{a}{c}+\frac{b}{c}=\frac{a+b}{c}$$ For example, you know that $$\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2}$$ and that $$\frac{1}{4}+\frac{1}{4}\neq\frac{1}{4}.$$

The correct form of the equation should be $$2\cos^2x-\sin x-1=0.$$ You will want to use the fact that for any $x$, $$\sin^2(x)+\cos^2(x)=1.$$

Thus, any occurrence of $\cos^2(x)$ can be replaced with $(1-\sin^2(x))$. $$2(1-\sin^2(x))-\sin(x)-1=0$$ $$[2\cdot1-2\cdot\sin^2(x)]-\sin(x)-1=0$$ $$-2\cdot\sin^2(x)+(-1)\sin(x)-1+2=0$$ $$(-2)\sin^2(x)+(-1)\sin(x)+(1)=0$$

Then, you will need to use the quadratic formula to solve for $\sin(x)$. It may help to write $y=\sin(x)$ temporarily, to prevent confusion. Keep in mind that there may be two different possible values of $\sin(x)$ that result.

$$(-2)\sin^2(x)+(-1)\sin(x)+(1)=0$$ $$(-2)y^2+(-1)y+(1)=0$$ $$y=\frac{-(-1)\pm\sqrt{(-1)^2-4(-2)(1)}}{2(-2)}=\frac{1\pm\sqrt{1+8}}{-4}=\frac{1\pm 3}{-4}=\begin{cases}\frac{1+3}{-4}=\frac{4}{-4}=\fbox{-1} &\text{ or }\\\\ & \\ \frac{1-3}{-4}=\frac{-2}{-4}=\fbox{\frac{1}{2}}.\end{cases}$$

Lastly, you need to find those values of $x$ that produce that value (or those values) of $\sin(x)$. Keep in mind that the function $\sin(x)$ is periodic; here is part of its graph to illustrate:
(and it continues similarly out to infinity in each direction) so that there are always infinitely many values of $x$ that will produce a given value of $\sin(x)$.

For any real number $x$, the values of $$\sin(x+2\pi k)$$ are all identical for any integer $k$,. There are usually even more values you can plug into $\sin$ that will give the same number (look at the graph for an idea).

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I have fixed my mistake. Thank you for pointing that out. 48 hours of no sleep is getting to me. So excuse my mistakes. –  Austin Broussard Jul 17 '12 at 3:36
@AustinBroussard For your own health, go and sleep. –  Pedro Tamaroff Jul 17 '12 at 3:38
I will after I finish a few more problems! –  Austin Broussard Jul 17 '12 at 3:41
We have different answers! Darn.. –  Austin Broussard Jul 17 '12 at 3:58
I too hastily approved of your equation $$2\cos^2(x)-\sin(x)+1=0$$ In fact, the correct equation is $$2\cos^2(x)-\sin(x)-1=0$$ When you had $$2\cos^2(x)=\sin(x)+1$$ you neglected to distribute the negative sign: $$2\cos^2(x)-[\sin(x)+1]=0$$ $$2\cos^2(x)-\sin(x)-1=0$$. –  Zev Chonoles Jul 17 '12 at 4:00

Hint 1: For all $x$, $\sin^2x+\cos^2x=1$
Hint 2: For $at^2+bt+c=0$ the solutions are $t_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

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Of course, $x$ cannot be an integer multiple of $\pi$ plus $\frac{\pi}{2}$. We have here a quadratic equation after a little manipulation, which you, yourself, started.

$$\frac{\sin x}{\cos x}+\frac{1}{\cos x}=2\cos x$$ After multiplying by $\cos x$, using the identity $1=\sin^2 x+\cos^2 x$, and bringing everything to the left hand side: $$2\sin^2 x+\sin x-1=0$$ If we replace $y=\sin x$, then you will be able find a familiar expression for $y$: $$2y^2+y-1 = 0$$ Which gives, $y = \frac{1}{2}$ or $-1$. So, $\sin x$ is equal to either of those two values, which yields $x= 2n\pi+\frac{\pi}{6}$, $x=2n\pi+\frac{5\pi}{6}$ or $x=2n\pi+\frac{\pi}{2}$. However, we ruled out the latter solution initially.

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\begin{eqnarray} \left(\dfrac{\sin x}{\cos x}\right)+\left(\dfrac{1}{\cos x}\right)&=&2\cos x\\ \sin x + 1 &=& 2 \cos^2 x \\ \sin x + 1 &=& 2(1-\sin^2 x)\\ \end{eqnarray} Then $\sin x = -1$ or $\sin x = 1/2$ and the solutions are $-\pi + 2k \pi, \pi/6 + 2 k\pi$ and $5\pi/6 + 2k \pi$.

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Can you explain the $+2k \pi$ part please. –  Austin Broussard Jul 17 '12 at 4:33
Only for solutions out the $[-\pi,\pi]$ –  user29999 Jul 17 '12 at 4:36

This is more a comment than an answer, but it is an important comment, since it affects the list of solutions. We seem to have solutions that come from $\sin x=\frac{1}{2}$, and solutions that come from $\sin x=-1$.

No problem with the $\sin x=\frac{1}{2}$ stuff. For the record, this gives the solutions $x=\frac{\pi}{6}+2k\pi$ and $x=\frac{5\pi}{6}+2k\pi$, where $k$ ranges over the integers, positive, negative, and $0$.

However, there is an issue with $\sin x=-1$. For then $\cos x=0$, and neither $\tan x$ nor $\sec x$ is defined. When we multiply through by $\cos x$, we must remember that if $\cos x=0$ we are multiplying both sides by $0$, so we are not getting an equivalent equation. Thus the "solutions" $-\frac{\pi}{2}+2k\pi$ must be discarded.

Remark: If we look at the function $\frac{\sin x+1}{\cos x}$, it turns out that it has a removable singularity at $x=-\frac{\pi}{2}$, and can be made continuous there by defining it to be $0$. In that sense, $x=-\frac{\pi}{2}$ is a solution. However, I doubt it would be accepted as a solution in the context the OP is in.

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So, ultimately is $-\frac{\pi}{2}$ a real answer for the OP –  Austin Broussard Jul 17 '12 at 4:56
@AustinBroussard: If it is high school, or early university, I am pretty sure that it would not be accepted as an answer. After all, if you "plug in" trying to compute $\tan(-\pi/2)$, the calculator will say error and refuse to do anything more until you reset it. Certainly the process by which it was obtained is flawed. One could make the removable discontinuity argument, which I did not write out. You would have to explain that the functions on the left are not defined at $-\frac{\pi}{2}$, but $\dots$. The safe thing is to explain why $-\frac{\pi}{2}$ is no good ($\tan$ not defined.) –  André Nicolas Jul 17 '12 at 5:08