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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 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 2010-12-02

2 Answers 2

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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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    And then you do the same thing with the first equation to see what its smallest positive solution is.2010-12-02
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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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    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}$2010-12-02