Question

What does it mean when it says an interval of (0, 2pi)?

Answer 1

It's referring to something (e.g. the shape of the graph) from the point where x is 0 to the point where x is 2pi. In trigonometry, this would correspond to the interval from where x is 0 degrees to when x is 360 degrees.

Answer 2

Side note: the round brackets mean to exclude 0 and 2pi from whatever you're considering. You include them if the brackets are [square].

Answer 3

Sorry, they were square.

Answer 4

So you include 0 and 2pi in whatever you're doing for that interval.

Answer 5

CAn I give you the problem that I'm working on?

Answer 6

Sure

Answer 7

Find all solutions in the interval [0, 2π).

2 sin2x = sin x

Answer 8

2 sin^2x=sinx

Answer 9

Since the sine function is a cyclic function, it repeats itself forever. For example, where is sin x = 0? At ... -3pi, -2pi, -pi, 0, pi, 2pi, 3pi, etc., or (pi)n (n = integer). There is an infinitew number of solutions. When you are asked what is the solution of an equation in an interval, then all they are looking for are the angles in that interval that satisfy the equation. So, what are the solutions of sin x = 0 in the interval [0,pi]? The answer is 0 and pi onl, and not (pi)n, n is an integer.

Answer 10

\[2\sin^2 x=\sin x\]\[2 \sin^2 x-\sin x=0\]\[\sin x(2 \sin x-1)=0\]

The you have sinx = 0 and 2sinx -1 = 0. Use both to solve for x, and you have your solutions.

Answer 11

How would I solve sinx=0?

Answer 12

For what values of x is the sin 0? x = sin^-1 (0)