If sin2x=0 ⇔ x=(kpi)/2sin2x=0⇔x=kπ2, why isn't sin5x=0 ⇔ x=(kpi)/5sin5x=0⇔x=kπ5?

How is sin5x=0 ⇔ x=(2kpi)/5 V x=(pi+2kpi)/5sin5x=0⇔x=2kπ5Vx=π+2kπ5?

2 Answers
Ananda Dasgupta
Jun 3, 2018

sin(5x) = 0sin(5x)=0 does imply x = (k pi)/5x=kπ5 where k in ZZ

Explanation:

I am assuming that the V in the expression

sin5x=0 ⇔ x=(2kpi)/5 V x=(pi+2kpi)/5

stands for "or", i,e, what was meant was

sin5x=0 ⇔ x=(2kpi)/5 vv x=(pi+2kpi)/5

Now x=(2kpi)/5 means that x is an even multiple of pi/5, while x = (pi+2k pi)/5 = ((2k+1)pi)/5 means that x is an odd multiple of pi/5.

Hence

x=(2kpi)/5 vv x=(pi+2kpi)/5

means that x is either an even multiple of pi/5 or an odd multiple of pi/5 - which simply means that it is a multiple of pi/5

Nghi N.
Jun 3, 2018

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. 5x = 0 + 2kpi --> x = (2kpi)/5
b. 5x = pi + 2kpi --> x = (2k + 1)(pi)/5
c. 5x = 2pi + 2kpi --> x = (k + 1)(2pi)/5

In case k = 0, there are 3 different answers:
a. x = 0
b. x = (pi)/5
c. x = (2pi)/5
In case k = 1, there are 3 answers:
a. x = (2pi)/5
b. x = (3pi)/5
c. x = (4pi)/5

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