1. ## help

sin2x=cos3x, solve for x

2. Originally Posted by lexmex1546
sin2x=cos3x, solve for x
You can do this,
$\sin 2x-\cos 3x=0$
$\sin 2x-\sin \left( \frac{\pi}{2} -3x \right)=0$

Use identity,
$\sin x-\sin y=2\cos \left( \frac{x+y}{2} \right)\sin \left(\frac{x-y}{2} \right)$

Thus,
$2\cos \left( \frac{\pi}{4}-\frac{5x}{2} \right) \sin \left( \frac{x}{2}-\frac{\pi}{4} \right)=0$

Thus, you have a two factors equal to zero.

3. Originally Posted by ThePerfectHacker
You can do this,
$\sin 2x-\cos 3x=0$
$\sin 2x-\sin \left( \frac{\pi}{2} -3x \right)=0$

Use identity,
$\sin x-\sin y=2\cos \left( \frac{x+y}{2} \right)\sin \left(\frac{x-y}{2} \right)$

Thus,
$2\cos \left( \frac{\pi}{4}-\frac{5x}{2} \right) \sin \left( \frac{x}{2}-\frac{\pi}{4} \right)=0$

Thus, you have a two factors equal to zero.
Oh now that's just neat!

-Dan

4. Hello, lexmex1546!

Here's a "cute" way to approach this problem . . .

Solve for $x:\;\;\sin2x\:=\:\cos3x$

$\text{Since }\,\sin\theta = \frac{opp}{hyp}\, \text{ and } \,\cos\theta = \frac{adj}{hyp}$

. . we see that $2x$ and $3x$ are angles in the same right triangle . . .

Code:
                        *
*  *
*  3x *
*        *
*           *
*              *
* 2x              *
*  *  *  *  *  *  *  *

Therefore: . $2x + 3x \:=\:90^o\quad\Rightarrow\quad\boxed{x \,=\,18^o}$

Of course, this is not a complete solution . . .

5. Originally Posted by lexmex1546
sin2x=cos3x, solve for x
The hard way.

$sin(2x) = 2sin(x)cos(x)$

$cos(3x) = cos(2x + x) = cos(2x)cos(x) - sin(2x)sin(x)$

$cos(3x) = (1 - 2sin^2(x))cos(x) - (2sin(x)cos(x))sin(x)$

$cos(3x) = cos(x) - 4sin^2(x)cos(x)$

Thus:
$2sin(x)cos(x) = cos(x) - 4sin^2(x)cos(x)$

Pull everything to the LHS and factor the common cos(x):
$(cos(x))(4sin^2(x) + 2sin(x) - 1) = 0$

So either $cos(x) = 0$ or $4sin^2(x) + 2sin(x) - 1 = 0$

For the sin solution, use the quadratic formula to find:
$sin(x) = \frac{-1 \pm \sqrt{5}}{4}$

So $x = asn \left ( \frac{-1 \pm \sqrt{5}}{4} \right )$

So the whole solution set in the domain $[0, 2\pi)$ will be x = $\pi/2, 3\pi/2$ from the $cos(x) = 0$ solution and x = $asn \left ( \frac{-1 + \sqrt{5}}{4} \right ) = \frac{\pi}{10}, \pi - \frac{\pi}{10} = \frac{9\pi}{10}$ from the sine solution. These may be easily seen to be the same solutions as ThePerfectHacker gave you, only from a much messier derivation.

-Dan