Question

x-intercepts of g(x)=cos (x-π)

Answer 1

You COULD expand that. \(\cos(x-\pi) = \cos(x)\cos(\pi)+\sin(x)\sin(\pi) = -\cos(x)\)

That might help.

Answer 2

x-intercept is where the graph touches the "x" axis, notice here

what's the value of "y" at that point? well, notice, y = 0 and "x" is ... we dunno

so if "y=0" at the x-intercept point, then we can say that \(\bf g(x)= cos(x-\pi)\implies 0 = cos(x-\pi)\\ \quad \\
cos^{-1}(0) = cos^{-1}(cos(x-\pi))\\ \quad \\
\textit{recall that } \quad cos^{-1}(cos(\theta)) = \theta\quad thus\\ \quad \\
cos^{-1}(0) = cos^{-1}(cos(x-\pi))\implies cos^{-1}(0) = x-\pi\)

now.. at what is the angle between \(\bf (0, \pi\) that has a cosine of 0?

Answer 3

I'm pretty sure it has the same x intercepts as cosx such as pie/2 and 3pie/2

Answer 4

pie is for eating
pi is for mathematics and Greek spellers.

Answer 5

Well excuse me math bully

Answer 6

RavenLynette though I must admit, your version is more tasty =)

Answer 7

?? I don't see any bullying. Just information for future use.

Frankly, I dare you to show me where any bullying took place.

I do have to agree with jdoe0001, i suppose.

Answer 8

Thank you for correcting my spelling but what is the answer to the problem?

Answer 9

You had it. \(\dfrac{\pi}{2} + k\pi\) for any integer k.