+0

# Here the answer is .. a - 3 values

0
547
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+1832

Here the answer is .. a - 3 values

http://i59.tinypic.com/2rogm52.jpg

right ! ?

Jul 31, 2014

#1
+121085
+10

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.

Jul 31, 2014

#1
+121085
+10

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.

CPhill Jul 31, 2014
#2
+1832
0

Thanx  CPhill Now its very clear  !!

Jul 31, 2014