# Finding trig intersection point

• Mar 4th 2011, 03:41 PM
Kuma
Finding trig intersection point
Hi

quick & easy question:

6 cos 4x = 6 sin 8x
cos 4x = sin 8x
cos 4x = 2 sin 4x cos 4x
1/2 = sin 4x
arcsin(1/2) = 4x
x = arcsin(1/2)/4

are the above steps correct?
If so, how can i find x in terms of pi?

Thanks.
• Mar 4th 2011, 03:53 PM
e^(i*pi)
Not quite. Dividing through by $\cos(4x)$ is potentially dividing by 0 so instead move all terms onto one side and factor

$2\sin(4x)\cos(4x) - \cos(4x) = \cos(4x)(2\sin(4x)-1) = 0$

Either $\cos(4x) = 0$ (these are the solutions you miss if you divide) or $2\sin(4x)-1 = 0$

As for $\arcsin(0.5)$ recall your unit circle to see that $\arcsin(0.5) = \dfrac{\pi}{6}$