# help

• Nov 18th 2006, 02:26 PM
lexmex1546
help
sin2x=cos3x, solve for x
• Nov 18th 2006, 02:34 PM
ThePerfectHacker
Quote:

Originally Posted by lexmex1546
sin2x=cos3x, solve for x

You can do this,
$\displaystyle \sin 2x-\cos 3x=0$
$\displaystyle \sin 2x-\sin \left( \frac{\pi}{2} -3x \right)=0$

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

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

Thus, you have a two factors equal to zero.
• Nov 18th 2006, 02:36 PM
topsquark
Quote:

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

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

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

Thus, you have a two factors equal to zero.

Oh now that's just neat! :)

-Dan
• Nov 18th 2006, 03:02 PM
Soroban
Hello, lexmex1546!

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

Quote:

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

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

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

Code:

                        *                     *  *                   *  3x *               *        *             *          *         *              *       * 2x              *   *  *  *  *  *  *  *  *

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

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

• Nov 18th 2006, 05:55 PM
topsquark
Quote:

Originally Posted by lexmex1546
sin2x=cos3x, solve for x

The hard way.

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

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

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

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

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

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

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

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

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

So the whole solution set in the domain $\displaystyle [0, 2\pi)$ will be x = $\displaystyle \pi/2, 3\pi/2$ from the $\displaystyle cos(x) = 0$ solution and x = $\displaystyle asn \left ( \frac{-1 + \sqrt{5}}{4} \right ) = \frac{\pi}{10}, \pi - \frac{\pi}{10} = \frac{9\pi}{10}$ 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