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

Question

$(sin x + cos x) \cdot \sqrt{2} = tan x + cot x$ $(sinx+cosx)*sqrt(2)=tanx+cotx$
Solving this equation I got x= $\frac{π}{4} + n π$ $pi/4+npi$ , but my professor said the solution should be x= $\frac{π}{4} + 2 n π$ $pi/4+2npi$
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 $2 π$ $2pi$ .

The period for both tan x and cot x is $π$ $pi$ .

The LCM is $2 π$ $2pi$ .

So, this is the overall period for

$f (x) = \sqrt{2} (cos x + sin x) - (tan x + cot x)$ $f( x ) = sqrt2 ( cos x + sin x ) - ( tan x + cot x)$ .

You can verify that

$f (x + 2 π) = f (x)$ $f( x + 2pi ) = f( x )$ but

$f (x + π) = - \sqrt{2} (cos x + sin x) - (tan x + cot x) \neq f (x)$ $f( x + pi )= - sqrt2 ( cos x + sin x ) - ( tan x + cot x) ne f(x)$ .

So, the general solution is

$x = \frac{π}{4} + 2 n π, n = 0, \pm 1, \pm 2, \pm 3, . .$ $x = pi/4 + 2npi, n = 0, +-1, +-2, +-3, ..$

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.