If #sin2x=0 ⇔ x=(kpi)/2#, why isn't #sin5x=0 ⇔ x=(kpi)/5#?
How is #sin5x=0 ⇔ x=(2kpi)/5 V x=(pi+2kpi)/5# ?
How is
2 Answers
Explanation:
I am assuming that the V in the expression
stands for "or", i,e, what was meant was
Now
Hence
means that
There are 3 different solutions in case sin 5x = 0
Explanation:
In fact, in the case (sin 5x = 0), there are 3 different solutions:
a.
b.
c.
In case k = 0, there are 3 different answers:
a.
b.
c.
In case k = 1, there are 3 answers:
a.
b.
c.