# sin4x=cos3x

• Feb 11th 2008, 12:19 PM
weasley74
sin4x=cos3x
How do I find all the solutions to sin4x=cos3x, [0, 2pi[ ?
• Feb 11th 2008, 06:39 PM
Soroban
Hello, weasley74!

Quote:

Find solutions to: . $\sin4x\:=\:\cos3x\;\;\text{on }[0,\,2\pi]$
Code:

                    *                   * *                 * 3x*           c  *    *             *      * a           *        *         * 4x        *       * * * * * * * *               b

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

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

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

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

• Feb 12th 2008, 11:29 PM
mr fantastic
Quote:

Originally Posted by Soroban
Hello, weasley74!

Code:

                    *                   * *                 * 3x*           c  *    *             *      * a           *        *         * 4x        *       * * * * * * * *               b

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

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

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

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

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

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

So, within the given domain, $x = \frac{\pi}{2}$.
• Feb 12th 2008, 11:41 PM
mr fantastic
Quote:

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

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

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

Another way of doing the question (which is how I picked up the missing solution):

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

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

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

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