You must be referring to post..
http://www.jiskha.com/display.cgi?id=1262496706
Please give yourself a "nickname" other than "anonymous", it makes it easier to find you for subsequent replies.
for any sin(kx) or cos(kx) the period is 2pi/k or 360º/k
your equation contains sin(5x) and cos(5x), so they both have a period of 2pi/5 radians.
that is why I added k(2pi/5) to each answer.
BTW, for tan(kx) the period is pi/k or 180/kº
How do you know when to add k(2pi) or k(pi)? What's the difference?
Also, for this problem, why are you supposed to use k(2pi/5) for the solutions, assuming that there is no domain?
2sin^2(5x)+sin(5x)-1=0
Thank you!
3 answers
Thank you! I will remember to use a nickname in the future. What would the periods be for csc, sec, and cot if I had to solve for those?
In a followup to this, why isn't cos(theta)=0 not just pi/2 + pi*k? For other values of cos(theta) this makes sense as +2pi*k, but does this not work for cod(theta)=0?