Question

Find period, domain, and range of y= 3 csc (3x + pi) -2
y=3 csc 3 (x+ pi/3) -2 but I'm stuck from here.
This is review from last year, but I've totally forgotten everything about trig functions. Not looking for just answers but also explanations!

Answer 1

The period of y=csc(x) is 2pi.
So the period of y=csc(bx) is 2pi/|b|.

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y=sin(x) on [-2pi,2pi] looks like:

We can see the period is 2pi.
The reciprocal function of y=sin(x) is y=csc(x).
So noticed the zeros of y=sin(x) on [-2pi,2pi], they were at x=-2pi,-pi,0,pi,2pi
y=csc(x) will be undefined there.
Notice where sin(x)=1 on [-2pi,2pi], they were at x=-3pi/2,x=pi/2
well this is where csc(x)=1 also since the reciprocal of 1 is still 1.
Notice where sin(x)=-1 on [-2pi,2pi], they were at x=-pi/2,3pi/2
well this where csc(x)=-1 since the reciprocal of -1 is still -1.
The graph of y=csc(x) looks like:

you can also from this graph that the range is (-inf,-1] union [1,inf)
and the domain is everything but where we had the zeros fro sin(x) which was
npi where n is integer
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Anyways...we have
\[\csc(x) \le -1 \text{ or } \csc(x) \ge 1 \text{ from the range } \\ \text{ this holds for any input } \\ \csc(3x+\pi) \le -1 \text{ or } \csc(3x+\pi) \ge 1 \\ \text{ multiply both sides by 3 and then subtract both sides by 2 } \\ \text{ \to find range of } y=3\csc(3x+\pi)-2\]
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Now for the domain part...
You could solve the following equation to find out what numbers to exclude:
\[\sin(3x+\pi)=0\]