1. ## sin4x=cos3x

How do I find all the solutions to sin4x=cos3x, [0, 2pi[ ?

2. Hello, weasley74!

Find solutions to: .$\displaystyle \sin4x\:=\:\cos3x\;\;\text{on }[0,\,2\pi]$
Code:
                    *
* *
* 3x*
c  *     *
*       * a
*         *
* 4x        *
* * * * * * * *
b

We see that: .$\displaystyle \sin4x \:=\:\frac{a}{c} \:=\:\cos3x$

. . and that $\displaystyle 4x$ and $\displaystyle 3x$ are complementary.

In general: .$\displaystyle 4x + 3x \:=\:\frac{\pi}{2} + 2\pi n \quad\Rightarrow\quad x \:=\:\frac{\pi}{14} + \frac{2\pi}{7}n$

Therefore: .$\displaystyle x \;=\;\frac{\pi}{14},\;\frac{5\pi}{14},\;\frac{9\pi }{14},\;\frac{13\pi}{14},\;\frac{17\pi}{14},\;\fra c{21\pi}{14},\;\frac{25\pi}{14}$

3. Originally Posted by Soroban
Hello, weasley74!

Code:
                    *
* *
* 3x*
c  *     *
*       * a
*         *
* 4x        *
* * * * * * * *
b

We see that: .$\displaystyle \sin4x \:=\:\frac{a}{c} \:=\:\cos3x$

. . and that $\displaystyle 4x$ and $\displaystyle 3x$ are complementary.

In general: .$\displaystyle 4x + 3x \:=\:\frac{\pi}{2} + 2\pi n \quad\Rightarrow\quad x \:=\:\frac{\pi}{14} + \frac{2\pi}{7}n$

Therefore: .$\displaystyle x \;=\;\frac{\pi}{14},\;\frac{5\pi}{14},\;\frac{9\pi }{14},\;\frac{13\pi}{14},\;\frac{17\pi}{14},\;\fra c{21\pi}{14},\;\frac{25\pi}{14}$

There's one more, I'm sorry to say ......

$\displaystyle x = \frac{\pi}{2} + 2 n \pi$.

So, within the given domain, $\displaystyle x = \frac{\pi}{2}$.

4. Originally Posted by mr fantastic
There's one more, I'm sorry to say ......

$\displaystyle x = \frac{\pi}{2} + 2 n \pi$.

So, within the given domain, $\displaystyle x = \frac{\pi}{2}$.
Another way of doing the question (which is how I picked up the missing solution):

Since $\displaystyle \sin A = \cos\left( \frac{\pi}{2} - A \right)$ (supplementary angle formula), the equation can be re-written as: $\displaystyle \, \cos\left( \frac{\pi}{2} - 4x \right) = \cos(3x)$.

Case 1: $\displaystyle \, \frac{\pi}{2} - 4x = 3x + 2m \pi \Rightarrow 7x = \frac{\pi}{2} + 2n \pi$ (where n = -m) and you have the solutions given by Soroban. Note that these solutions include $\displaystyle \frac{21 \pi}{14} = \frac{3 \pi}{2}$.

Case 2: Since $\displaystyle \, \cos (-\theta) = \cos(\theta)$ it can also be said that

$\displaystyle - \left( \frac{\pi}{2} - 4x \right) = 3x + 2n \pi \Rightarrow \Rightarrow -\frac{\pi}{2} + 4x = 3x + 2n \pi \Rightarrow x = \frac{\pi}{2} + 2n \pi$.

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# sin4x-cos3x=0

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