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How to find the least positive root of the equation $\cos 3x + \sin 5x = 0$?

My approach so far is to represent $\sin 5x$ as $\cos \biggl(\frac{\pi}{2} - 5x\biggr)$ then the whole equation reduces to $$2\cos \biggl(\frac{\pi}{4} - x\biggr)\cdot \cos \biggl(\frac{\pi}{4} - 4x\biggr) = 0$$

From here we can write:

$$\biggl(\frac{\pi}{4} - x\biggr) = n\pi + \frac{\pi}{2} , n \in \mathbb{Z}$$

$$\biggl(\frac{\pi}{4} - 4x\biggr) = n\pi + \frac{\pi}{2} , n \in \mathbb{Z}$$

Now there can be infinitely many solutions for this, what I am not getting how to compute the minimum among them? And what about if I am asked to find the maximum?

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The solutions of your last two equations are $x=-\frac{1}{4}\pi -n\pi $, $n\in \mathbb{Z}$ and $x=-\frac{1}{16}\pi -\frac{1}{4}n\pi $, $n\in \mathbb{Z}$ –  Américo Tavares Dec 2 '10 at 10:29
    
The first positive solution of the form $x=-\frac{1}{4}\pi -n\pi $ is $-\frac{1}{4}\pi +\pi =\frac{3}{4}\pi $, while of the form $-\frac{1}{16}\pi -\frac{1}{4}\pi n$ is $-\frac{1}{16}\pi +\frac{1}{4}\pi =\frac{3}{16}\pi <% \frac{3}{4}\pi $ –  Américo Tavares Dec 2 '10 at 10:42

2 Answers 2

up vote 5 down vote accepted

You've already done the difficult part! Now just find all the solutions of $\cos(\pi/4-x)=0$ and $\cos(\pi/4-4x)=0$, and pick the smallest positive one.

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Hmm.. but the problem is when I am equating them I am not getting the correct answer which must be $\frac{3\pi}{16}$ –  Quixotic Dec 2 '10 at 9:56
    
So I guess I have some problems in equating them, what I am doing is writing $\pi/4-x = \pi/2$ and then finding $x$ and the same thing for the other one too. –  Quixotic Dec 2 '10 at 9:58
    
If you do that, you're not finding all the solutions. Remember that $\cos a=\cos b$ iff $a=\pm b + 2\pi n$ where $n$ is an arbitrary integer. –  Hans Lundmark Dec 2 '10 at 10:03
    
How exactly I can find all the solutions? there are infinitely large number of solutions. –  Quixotic Dec 2 '10 at 10:05
1  
Oh, I didn't see your edit. The second of your equations can be solved for $x$. This gives $x=-\pi/16 - n \pi/4$. If you take $n=0$ you get a negative $x$, and clearly also if you take $n>0$. But if you take $n=-1$ then you get $x=-\pi/16+\pi=3\pi/16$ which is positive, and it's the smallest positive solution since other negative values of $n$ will give you larger solutions. (There is no largest solution, though. Taking $|n|$ big enough you can find as large solutions $x$ as you wish.) –  Hans Lundmark Dec 2 '10 at 10:24

EDIT: So apparently you want a full solution...

A product of two real numbers is $0$ precisely when a factor is $0$, so you must have that $$ \begin{align} & \cos\left(\tfrac{\pi}{4} - x\right) = 0 \qquad\;\; (1) \text{, or} \\\\ & \cos\left(\tfrac{\pi}{4} - 4x\right) = 0 \qquad (2). \end{align} $$ Now we use that $\cos a = 0 \iff a \in \tfrac{\pi}{2} + \pi\mathbb{Z}$: $$ \begin{align} \cos \left(\tfrac{\pi}{4} - x\right) = 0 & \iff \tfrac{\pi}{4} - x \in \tfrac{\pi}{2} + \pi\mathbb{Z} \\\\ & \iff x \in \tfrac{3\pi}{4} + \pi \mathbb{Z}, \end{align} $$ the smallest positive $x$ satisfying this clearly being $\tfrac{3\pi}{4}$. $$ \begin{align} \cos \left(\tfrac{\pi}{4} - 4x\right) = 0 & \iff \tfrac{\pi}{4} - 4x \in \tfrac{\pi}{2} + \pi\mathbb{Z} \\\\ & \iff 4x \in \tfrac{3\pi}{4} + \pi \mathbb{Z} \\\\ & \iff x \in \tfrac{3\pi}{16} + \tfrac{\pi}{4}\mathbb{Z}, \end{align} $$ the smallest positive $x$ satisfying this clearly being $\tfrac{3\pi}{16}$. The problem is now reduced to picking the smallest of two real numbers.

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How exactly I am suppose to pick up the smallest real? –  Quixotic Dec 2 '10 at 10:06
    
No,it's not the full solution, what I am not getting how are you writting $\frac{\pi}{4} - x \in \frac{\pi}{2} + \pi\mathbb{Z} \iff x \in \tfrac{3\pi}{4} + \pi \mathbb{Z}$ i.e the change of sighn here,it may be stupid but I could not understand! –  Quixotic Dec 2 '10 at 10:24
    
That bit is fairly simple: $\tfrac{\pi}{4}-x \in \tfrac{\pi}{2} + \pi\mathbb{Z} \iff -x \in \tfrac{\pi}{4} + \pi\mathbb{Z} \iff x \in -\tfrac{\pi}{4} + \pi\mathbb{Z} = \tfrac{3\pi}{4} + \pi\mathbb{Z}$ –  kahen Dec 2 '10 at 10:30

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