Hello Paymemoney 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?
Grandad