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

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- Jan 7th 2010, 02:07 AMPaymemoneySolving 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 AMRaoh
- Jan 7th 2010, 02:38 AMGrandad
Hello PaymemoneyNoting 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:$\displaystyle 2\sin x \cos x = \cos x$Can you go on to complete this now?

$\displaystyle \Rightarrow \cos x = 0$ or $\displaystyle 2\sin x = 1$

__Spoiler__:

Grandad - Jan 7th 2010, 02:45 AMPaymemoney
thanks i get how to do it now