5sin(kx) = 4

acemi123

New member
Joined
Apr 14, 2019
Messages
31
which one is correct?

a) only if k < 2pi
b) all real number for k
c) only if -1<= k <=1
d) only if k>0


After
sin(kx) = 4/5
I can't continue.
I can guess it's positiv which are first and second quadrant. and I think it cant be 0 because equation =4/5. if sin(0*x)=0
but after all i couldnt figure ou to solve.
Help please.
 
I'm not sure what the question means, as you have translated it. Is it asking, under which of the four conditions does a solution exist? Certainly none of the choices listed is the actual solution, since they don't involve x.

Certainly you can solve the equation 5 sin(kx) = 4 by continuing from where you left off, given any nonzero value of k. But that doesn't fit any of the choices.

If necessary, can you show us the actual problem in the original language, so we can check the translation ourselves?
 
I'm not sure what the question means, as you have translated it. Is it asking, under which of the four conditions does a solution exist? Certainly none of the choices listed is the actual solution, since they don't involve x.

Certainly you can solve the equation 5 sin(kx) = 4 by continuing from where you left off, given any nonzero value of k. But that doesn't fit any of the choices.

If necessary, can you show us the actual problem in the original language, so we can check the translation ourselves?
here oroginal questin.

12918
 
It looks like my guess was right; Google translates it as

The equation 5 sin (kx) = 4 has solution (a) only if k <2 pi; (b) for each value of k in reals; ...​

Unfortunately, my answer is still "for any real value of k except 0", just as you suggested. It is not necessary for k to be positive, so (d) is not correct.
 
Top