+1832 Here the answer is .. a - 3 values
http://i59.tinypic.com/2rogm52.jpg
But if he asks about cosine the answer will be 2 values !
right ! ?
+128475 sin (2Θ) =0 .....I'll use "Θ" instead of "x"........it's arbitrary, anyway......
The sine is 0 at 0, pi and 2pi. So, since we have 2Θ, we need to divide each of these by 2 and we get 0, pi/2 and pi.
Note that, the next value where the sine = 0 would be at 3pi. And dividing this by 2, we get 3pi/2.
And the next value where sine = 0 would be at 4pi. And dividing this by 2, we get 2pi. And we're out of values.
So, there are 5 values of x in [0, 2pi] that satisfy this question.....here's a graph of sin(2Θ) on the requested interval...
This makes sense, because the "2Θ" "compresses" the normal period. Thus, in 2pi, there are two periods, instead of one.
+128475 sin (2Θ) =0 .....I'll use "Θ" instead of "x"........it's arbitrary, anyway......
The sine is 0 at 0, pi and 2pi. So, since we have 2Θ, we need to divide each of these by 2 and we get 0, pi/2 and pi.
Note that, the next value where the sine = 0 would be at 3pi. And dividing this by 2, we get 3pi/2.
And the next value where sine = 0 would be at 4pi. And dividing this by 2, we get 2pi. And we're out of values.
So, there are 5 values of x in [0, 2pi] that satisfy this question.....here's a graph of sin(2Θ) on the requested interval...
This makes sense, because the "2Θ" "compresses" the normal period. Thus, in 2pi, there are two periods, instead of one.
