OpenStudy (anonymous):

solve cos (7x)=0 on the interval [0,2pi/7]

OpenStudy (anonymous):

Let's replace \(7x\) with another variable -- how about \(u\)? Can you solve \(\cos u=0\) on the interval \([0,2\pi]\)?

OpenStudy (anonymous):

Phew thanks for the reply...So cosx=0 at 0 and 2pi right??

OpenStudy (anonymous):

No... try sketching a unit circle.

OpenStudy (anonymous):

oh its at pi/2, 3pi/2...

OpenStudy (anonymous):

@tanentrahan indeed. Now convert back to \(x\). Remember, \(u=7x\) so \(x=\frac17u\). Since we solved for \(u\) in \([0,2\pi)\) we will get \(x\) in \(\left[\frac07,\frac{2\pi}7\right)=\left[0,\frac{2\pi}7\right)\).

OpenStudy (anonymous):

Thanks!!! I'm not sure if i completely understand what happened though. We isolated the "x" and solved for it right??

OpenStudy (anonymous):

So for this one cos(7x)=-1 the interval being 0, pi/2 our x value is -1/7 right?