How can I check if the period of a trigonometric equation solution is #npi" or "2npi# ?

Question

#(sinx+cosx)*sqrt(2)=tanx+cotx#
Solving this equation I got x= #pi/4+npi# , but my professor said the solution should be x= #pi/4+2npi#
How is this checked?

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Answer 1 · 2018-07-08T14:32:51

See the explanation, and graph for verification.

The least positive P for which f (x + P) = f (x ) is the period of f( x ).

The period for both sin x and cos x is #2pi#.

The period for both tan x and cot x is #pi#.

The LCM is #2pi #.

So, this is the overall period for

#f( x ) = sqrt2 ( cos x + sin x ) - ( tan x + cot x)#.

You can verify that

#f( x + 2pi ) = f( x )# but

#f( x + pi )= - sqrt2 ( cos x + sin x ) - ( tan x + cot x) ne f(x)#.

So, the general solution is

#x = pi/4 + 2npi, n = 0, +-1, +-2, +-3, ..#

Graph reveals this.
graph{y - sqrt 2 (sin x + cos x )+tan x + cot x = 0[-20 20 -10 10]}

The graph braces x-axis, for the solutions, at

#x = ... 18.6, -11.8, -5.5, 0.79, 7.07, 13.4 ...# for

#= ...- (23/4)pi, -(15/4)pi, -(7/4)pi, pi/4, (9/4)pi, (17/4)pi...,#

respectively, as approximations.