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#?

2 Answers
Ananda Dasgupta
Jun 3, 2018

#sin(5x) = 0# does imply #x = (k pi)/5# 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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