# Solving trig

• Jan 7th 2010, 02:07 AM
Paymemoney
Solving trig
Need help on how solve the following equation for 0 <= x <= $\displaystyle 2\pi$:
This is what i have done, but it is incorrect.

$\displaystyle sin(2x)=cos(x)$
$\displaystyle \frac{sin(2x)}{cos(x)}=0$
$\displaystyle tan2x=0$which gets 0

P.S
• Jan 7th 2010, 02:22 AM
Raoh
Quote:

Originally Posted by Paymemoney
Need help on how solve the following equation for 0 <= x <= $\displaystyle 2\pi$:
This is what i have done, but it is incorrect.

$\displaystyle sin(2x)=cos(x)$
$\displaystyle \frac{sin(2x)}{cos(x)}=0$
$\displaystyle tan2x=0$which gets 0

P.S

your second step is wrong and so what follows.
(Happy)
• Jan 7th 2010, 02:38 AM
Hello Paymemoney
Quote:

Originally Posted by Paymemoney
Need help on how solve the following equation for 0 <= x <= $\displaystyle 2\pi$:
This is what i have done, but it is incorrect.

$\displaystyle sin(2x)=cos(x)$
$\displaystyle \frac{sin(2x)}{cos(x)}=0$
$\displaystyle tan2x=0$which gets 0

P.S

Noting that $\displaystyle \sin2x = 2\sin x \cos x$, we get:
$\displaystyle 2\sin x \cos x = \cos x$
You need to be a little careful now. Don't just divide both sides by $\displaystyle \cos x$, because it might be zero, and the first commandment of arithmetic is 'Thou shalt not divide by zero'. So you need to say:
$\displaystyle 2\sin x \cos x = \cos x$

$\displaystyle \Rightarrow \cos x = 0$ or $\displaystyle 2\sin x = 1$
Can you go on to complete this now?
Spoiler:
(I get the answers $\displaystyle x = \frac{\pi}{6},x = \frac{\pi}{2},x = \frac{5\pi}{6},x = \frac{3\pi}{2}$.)