Pre-Calc help needed now!!! Please help mee!!?

Question

1. The number of solutions to the equation sin (2000x) = 3/7 in the interval [0, 2π ] is

A. 1000
B. 2000
C. 4000
D. 6000
E. 8000

2. If x=a is an asymptote of the secant function, then cot(a) = 0.
True or false? and why?

Answer 1

TRUE
If x = a is an asymptote, then the denominator is zero.
sec(x) = 1/ cos(x) ===> which means cos(a) = 0

Now, cot(x) is defined cos(x) / sin(x) ===> from above, cos(a)=0
then cot (a) = 0 / sin(a) = 0

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since 3/7 is between 0 and 1, the equation sin x = 3/7 will usually have 2 solutions in the interval.
We have changed the period from 2pi ===> to 2pi/2000
or horizontally shrunk the graph by 2000.
so we know have 2000 times as many = 2000*2 = 4000

C. 4000 solutions in[0, 2pi]

Answer 2

1 C.
2 This is true. If the secant function is approaching an asymptote, it means its value is approaching infinity (either + or minus) Thus, the cosine function --> 0. Since cot(a) = cos(a)/sin(a), cot(a) --> 0 also