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August 30, 2014

August 30, 2014

Posted by **Anonymous** on Sunday, January 3, 2010 at 1:39pm.

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!

- Trigonometry -
**Reiny**, Sunday, January 3, 2010 at 1:56pmYou 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

- Trigonometry -
**Anonymous**, Sunday, January 3, 2010 at 3:26pmThank 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?

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