At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

This is a sine wave shifted down 3 units, with an amplitude five times that of the parent function, and a period of 4. Thus the equation of the function is f(x)=5 sin( (pi*x)/4)-3
0=5 sin( (pi*x)/4)-3
Then, the x coordinate is 3/5=sin((pi/4)x)
(pi*x)/4=arcsin(3/5)
X=(4arcsin(3/5))/pi
My work. The last two lines are wrong because of the arcsin. Please explain to me how to determine the second intersection p and not the first intersection.

what did you get for x?

I'm really just having trouble understanding how to find each solution in this problem, and not just the 0

So I take advantage of the fact that the sine graph is symmetrical?

Yes.

Uh, what about trig equations?

for 3c I have
\(\large \cos^{-1}(\cos(x))=\pi/3\)

I mean I'm tempted to just cancel out the functions, but I know that's wrong.

Wait but since we take the arccos second, we can cancel them out right?

Even then, I don't think you can cancel them unless you're in a very restricted domain.

According to wolfram, that only works for \(-1\le x\le1\)