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!

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

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?

When solving trigonometric equations like these, the variables k(2pi) and k(pi) are introduced to represent multiple solutions due to the periodic nature of trigonometric functions.

The difference between k(2pi) and k(pi) lies in the number of solutions you are considering.

- If you use k(2pi), you are accounting for all possible solutions within one full revolution or 360 degrees. In other words, k(2pi) represents the full period of the trigonometric function. This is typically used when you want to find all solutions without any restrictions on the domain.

- On the other hand, if you use k(pi), you are considering only half of the range, which is half a period or 180 degrees. This is typically used when there are some domain restrictions or if you want to focus on a specific range of solutions.

Now coming to the problem you mentioned, 2sin^2(5x) + sin(5x) - 1 = 0. In order to find the solutions, you can follow these steps:

Step 1: Notice that we have a quadratic trigonometric equation in terms of sin(5x). To make it simpler, let's introduce a substitution u = sin(5x). Now the equation becomes 2u^2 + u - 1 = 0.

Step 2: Solve the quadratic equation 2u^2 + u - 1 = 0 using factoring, completing the square, or the quadratic formula. The solutions for u will give you possible values for sin(5x).

Step 3: Once you find the solutions for u, set each solution equal to sin(5x) and solve for x. This will give you the values for x.

Step 4: Now, since we have sin(5x) as the variable, we need to determine the range of solutions we are looking for. If no domain restrictions are specified, you can use the full period k(2pi) for the solutions. Therefore, you will use k(2pi/5) for the solutions.

By using k(2pi/5) as the general solution, you will find all the solutions within the given domain.

Remember, in general, the choice between k(pi) and k(2pi) depends on the specific problem and the context of the question. Always check if there are any domain restrictions or if the problem specifies a certain range for the solutions.