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solve cos (7x)=0 on the interval [0,2pi/7]
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Let's replace \(7x\) with another variable -- how about \(u\)? Can you solve \(\cos u=0\) on the interval \([0,2\pi]\)?
Phew thanks for the reply...So cosx=0 at 0 and 2pi right??
No... try sketching a unit circle.
oh its at pi/2, 3pi/2...
@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 \(\left[\frac07,\frac{2\pi}7\right)=\left[0,\frac{2\pi}7\right)\).
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Thanks!!! I'm not sure if i completely understand what happened though. We isolated the "x" and solved for it right??
So for this one cos(7x)=-1 the interval being 0, pi/2 our x value is -1/7 right?
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